Powers

πŸ†Practice exponents rules

What is an exponent?

Powers are the number that is multiplied by itself several times.
Each power consists of two main parts:Β 

  • Base of the power: The number that fulfills the requirement of duplication. The principal number is written in large size.
  • Exponent: the number that determines how many times the power base needs to be multiplied by itself.
    The exponent is written in small size and appears on the right side above the power base.

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Test yourself on exponents rules!

einstein

\( 112^0=\text{?} \)

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What is necessary to know about exponents?

Powers have rules and norms with which we must be familiar.
Click on each of the rules to read in detail.

  • Multiplication of Powers with the Same Base amΓ—an=a(m+n)a^m\times a^n=a^{(m+n)}
  • Quotient of Powers with the Same Base aman=a(mβˆ’n)\frac {a^m}{a^n} =a^{(m-n)}
  • Power of a Product (aΓ—b)n=anΓ—bn(a\times b)^n=a^n\times b^n
  • Power of a Quotient (ab)n=anbn(\frac {a}{b})^n=\frac {a^n}{b^n}
  • Power of a Power (an)m=a(nΓ—m)(a^n )^m=a^{(n\times m)}
  • Power with Exponent 0 a0=1a^0=1
  • Powers with a Negative Integer Exponent aβˆ’n=1ana^{-n}=\frac {1}{a^n}

Decomposing Natural Numbers into Multiplication Powers

We can use powers and their rules to break down natural numbers by multiplying the powers of prime numbers.
We will draw a line next to the number we want to divide and try to divide it by the smallest number we can. If the number is divisible without any remainder, we will write the prime factor in front of our number and below our number the result we got when we divided it by the prime factor.
We will continue breaking down the number in the same way until we reach the number 1 1 in the left column which cannot be broken down any further.
The multiplication of the factors is the multiplication among the factors we mentioned on the left and is equal to the natural number that we decomposed.


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Exponential Equations

Exponential equations are equations in which the unknown appears as an exponent.
These equations can be solved in two ways:Β 

Identical Bases
Reaching the state where the bases are the same.
If the bases are equal, we can compare the equations.
In general:
When
ax=aya^x=a^y
Then
x=yx=y

Quadratic Equation and Substitution of (t t )

When we want to solve exponential equations using a quadratic equation, we need to bring our exercise to a state where there is one element, the same squared, and a free number (without unknowns), for example: (4x)2+4Γ—4xβˆ’12=0 (4^x)^2+4\times4^x-12=0 .
Once we have reached this state, we will place t in place of the element we have (the entire element - including its power), in our case: t=4xt=4^x
Thus, we will arrive at an ordinary quadratic equation that we can easily solve, with our example: t2+4tβˆ’12=0t^2+4t-12=0
We must not forget to substitute the values of t t we found back into the corresponding equation to find the values of the unknown we are looking for X X .


Scientific Notation

Another use of exponents is the ability to express very large or very small numbers in a convenient way for reading and writing.
In scientific notation, the number is displayed as the multiplication of a number that is between 1 1 and 10 10 by 10 10 raised to some power.
Generally, the expression will look like this:

mΓ—10e m\times10^e

m m will be a number between 0 0 and 1 1 .
If e is a positive integer, the entire expression will be a number greater than 1 1 .
If e is a negative integer, the entire expression will be a number less than 1 1 .


Think of an exponent as a requirement for the number it operates on.
Exponentiation means describing how many times the number is required to multiply by itself.
For example, if we are given a 3 3 raised to the 5 5 , that is: 353^5
We can say that the number 3 3 is required to multiply by itself 5 5 times, as its power number.
Therefore:
35=3Γ—3Γ—3Γ—3Γ—33^5=3\times 3\times 3\times 3\times 3


Do you know what the answer is?

Exponentiation

So, what are powers really?

Base of the power, (in black) is the main number on which the operation is performed. In this example, the base of the power is 3 3 .
The exponent, (in orange) is the number that indicates how many times the main number should be multiplied by itself.
The exponent appears small to the right, above the base of the power. In this example, the exponent is 5 5 .

Powers have a set of rules that are very important to know in depth in order to solve power exercises quickly and without making mistakes.
Don't worry, Tutorela brings together everything you need to know about powers.
So, shall we get started?

Let's learn the first rule about powers:


Rules of Exponentiation

Let's learn about the first rule of exponents:

Multiplication of Powers with the Same Base

When we are given an exercise with a multiplication operation between identical bases that each have a different exponent, we can add the exponents to reach a state where we have a single base with one exponent.
Let's see this in the example:
42Γ—43= 4^2\times4^3=

In this exercise, we can identify that the bases are the same (both bases are 4 4 ) and there is a multiplication operation between them.
Therefore, we can add the exponents to obtain:

43+24^{3+2}
That is:
45=10244^5=1024

Let's explain the logic behind this rule:
For example:Β 

Please note that we obtained the same result.
When there is a multiplication operation between identical bases and they are powers, we can add the exponents. The number of exponents we obtain will become the new exponent of the same base in the exercise.
This rule also works in algebraic expressions as follows:
x2Γ—x2Γ—x3=x^2\times x^2\times x^3=
Since the base is the same, and there is a multiplication operation between the same bases, we can add the exponents and obtain:
x2+2+3=x^{2+2+3}=
x7x^7

Let's apply the rule of multiplying powers with the same base in a formula:
amΓ—an=a(m+n) a^m\times a^n=a^{(m+n)}


Check your understanding

A Quotient of Powers with the Same Base

When we are given a quotient or an exercise with a division operation between identical bases, we can subtract the exponents in the order they appear to obtain a single base with a single power.
Let's demonstrate this with an example:
2523=\frac{2^5}{2^3} =
In this exercise, we can identify that the bases are the same (both bases are 2 2 ) and that there is a division operation between them.
Therefore, we can subtract the exponents and obtain:
25βˆ’3=222^{5-3}=2^2


Pay attention!
The subtraction of powers will be done in the order they appear in the exercise. We subtract the power in the numerator, then in the denominator, and not the other way around.

Let's explain the logic behind this rule:

If we take the exercise

2523=\frac{2^5}{2^3} =

And try to simplify it to see what it really tells us we will obtain:
2Γ—2Γ—2Γ—2Γ—22Γ—2Γ—2= \frac{2\times2\times2\times2\times2}{2\times2\times2}=

We can reduce and reach

Please note that we obtained the same result.
When there is a division operation (in the form of a fraction or another way) between identical bases and they are powers, we can subtract the exponents. The difference in powers that we obtain will become the new exponent of the same base in the exercise.
This rule also works in algebraic expressions as follows:Β 

x4x2=\frac{x^4}{x^2} =

Since the bases are the same, and there is a division operation between them, we can subtract the exponents and obtain:

x4βˆ’2=x2 x^{4-2}=x^2

Let's apply the rule of the quotient of powers with equal bases in a formula:

aman=a(mβˆ’n)\frac {a^m}{a^n} =a^{(m-n)}

We will move on to the third rule of powers:


Power of a Product

When we have a power that is applied to an entire product or an entire exercise where there are only multiplication operations, we can take the exponent and apply it to each of the elements.
We will leave the multiplication operation between the elements.

(5Γ—3)3=(5\times 3)^3=
We can see that the power 33 applies to the entire product, the entire exercise. Therefore, we can take the exponent and apply it to each of the elements. Don't forget, leave the multiplication operation between the elements.
We will obtain:

53Γ—33=33755^3\times 3^3=3375
We can use this rule even if there are more than two elements in the exercise, as long as all the operations between the bases are multiplications.

This rule also works in algebraic expressions as follows:Β 
(xΓ—2xΓ—4x)2=(x\times 2x\times 4x)^2=
x2Γ—(2x)2Γ—(4x)2=x^2\times (2x)^2\times (4x)^2=
Pay attention, we can also apply the power of a product rule to 2x2x
Or to 4x4x . As there is also a multiplication operation between the coefficient for xx.
Therefore, we will apply the power of a product rule again and obtain:
x2Γ—22Γ—x2Γ—42Γ—x2=x^2\times 2^2\times x^2\times 4^2\times x^2=
x2Γ—4Γ—x2Γ—16Γ—x2=x^2\times 4\times x^2\times 16\times x^2=
Now, we can use the first rule we learned: multiplying powers with the same bases and combining the powers of the same bases.
We will obtain:
x6Γ—64=64x6x^6\times 64=64x^6
Let's apply the power of a product rule to a formula:

(aΓ—b)n=anΓ—bn(a\times b)^n=a^n\times b^n
Let's move on to the fourth power rule:


Do you think you will be able to solve it?

Power of a Quotient

When we have a power that is applied to the entire quotient or the entire exercise where there are only division operations, we can take the exponent and apply it to each of the elements.
Between the elements, we will leave the division operation - the fraction line.
Let's see this in an example:
(23)4=(\frac {2}{3})^4=
We can see that the power 44 is over the entire quotient, over the entire exercise. Therefore, we can take the exponent and apply it to each of the elements. We will not forget to leave the division operation - the fraction line, between the elements.
We will obtain:

2434=1681\frac {2^4}{3^4}=\frac{16}{81}

This rule also works in algebraic expressions as follows:Β 

(x2x)3=( \frac {x}{2x})^3=

x323Γ—x3= \frac{x^3}{2^3\times x^3}=
We simplify and obtain: Β 
181 \over 8

We apply the power of a quotient rule in a formula:

(ab)n=anbn(\frac {a}{b})^n=\frac {a^n}{b^n}

Let's move on to the fifth power rule:


Power of a Power

When we encounter an exercise with a power base with an exponent and another power on top (using parentheses), we can double the powers and convert the product we obtained into a new power that will be the power base.
It might sound a bit complicated, but it's not at all.
Let's see an example:
(43)2=(4^3 )^2=
We can see that inside the parentheses there is a base with an exponent. Outside the parentheses, there is another power that influences everything inside the parentheses.
According to the rule of power of a power, we can multiply the exponents and the result we get is operated as the exponent in function of the existing power.
Therefore, we will obtain:
43Γ—2=4^{3\times 2}=
46=40964^6=4096

Let's explain the logic behind this rule:
If we simplify the exercise
(43)2=(4^3 )^2=
It seems that we understand exactly what it is telling us, we will get:

(4Γ—4Γ—4)2=(4\times 4\times 4)^2=
According to the rule of power of a product that we have already learned, we will get:

42Γ—42Γ—42=4^2\times 4^2\times 4^2=
Now, we can use the first rule we learned, multiply powers with the same bases and add all the powers. We will obtain:
46=40964^6=4096
Note that we arrived at the same result.

This rule also works in algebraic expressions as follows:
(x2)3=(x^2 )^3=
x2Γ—3=x^{2\times 3}=
x6x^6
Let's apply the power of a power rule to the formula:

(an)m=a(nΓ—m)(a^n )^m=a^{(n\times m)}

We move on to the sixth rule of powers:


Test your knowledge

Powers with Zero Exponent

When we have a number (power base) to which a power of 00 is applied, that is, the exponent is 00, we obtain 11 in any situation, except when the power base is 00.
In fact, as long as the power base is not 00, any number raised to the power of 00 will be equal to 11.

Let's see this in the example:

50=15^0=1

We can see that the number 55, when raised to the power of 00, will therefore be equal to 11.
This rule also works in algebraic expressions.
We will see it as follows:
When x≠0x≠0
x0=1x^0=1

we apply the power rule with a zero exponent in the formula:

a0=1a^0=1
When a≠0a≠0

We move on to the seventh power rule:


Powers with Negative Integer Exponents

When we have an exercise in which the exponent is a negative integer, we will convert the expression into a fraction.
The numerator will be 11.
The denominator will be the base of the power with the positive exponent.
Let's see this in the example:
5βˆ’3=5^{-3}=
In this exercise, we can see that the exponent is a negative integer. Therefore, we will convert the expression to a fraction where the numerator is 11 and the denominator has a power base with the positive exponent.
We obtain:
153=1125\frac {1}{5^3}=\frac{1}{125}

In the same way, even if we have a negative power base, we can use it normally.
For example:

(βˆ’4)βˆ’5=1(βˆ’4)5=1βˆ’1024(-4)^{-5}=\frac {1}{(-4)^5}=\frac{1}{-1024}

Pay attention, convert the symbol only for the exponent and not for the power base.

This rule also works in algebraic expressions as follows:
xβˆ’3=1x3x^{-3}=\frac {1}{x^3}

We apply the rule of power with negative integer exponent in the formula:

aβˆ’n=1ana^{-n}=\frac {1}{a^n}
Use of every power rule
So far, we have studied each law separately. It is important that you know how to combine the power rules and use them in combination in a single exercise.
Do not worry, if you practice each rule separately and then move on to practice combining all the power rules, you will be able to easily solve an exercise that requires using all the power rules.
First, let's gather all the power rules we have learned, so we have them before our eyes:
amΓ—an=a(m+n)a^m\times a^n=a^{(m+n)}
aman=a(mβˆ’n)\frac {a^m}{a^n} =a^{(m-n)}
(aΓ—b)n=anΓ—bn(a\times b)^n=a^n\times b^n
(ab)n=anbn(\frac {a}{b})^n=\frac {a^n}{b^n}
(an)m=a(nΓ—m)(a^n )^m=a^{(n\times m)}
a0=1a^0=1
When a≠0a≠0
aβˆ’n=1ana^{-n}=\frac {1}{a^n}

We will solve the following exercise that requires the use of all the power rules:

(xy)βˆ’3Γ—(xβˆ’3)4Γ—yβˆ’2yβˆ’4=(\frac{x}{y})^{-3}\times \frac{\left(x^{-3}\right)^4\times y^{-2}}{y^{-4}}=
We know that this exercise can be very daunting, but trust us, it includes everything you have already learned.
The key to solving it is simply to decipher what you can do with the power rules you have learned. In the exercise, we will notice that it has elements with parentheses. According to the order of arithmetic operations, parentheses precede each one and, therefore, we will solve them first.
We will start with the first element that has parentheses: (xy)βˆ’3(\frac{x}{y})^{-3}
According to this rule:Β (ab)n=anbn(\frac {a}{b})^n=\frac {a^n}{b^n}

We obtain that:
(xy)βˆ’3=xβˆ’3yβˆ’3(\frac{x}{y})^{-3}=\frac{x^{-3}}{y^{-3}}
We move on to the second element that contains parentheses:
(xβˆ’3)4(x^{-3})^4

According to this rule: Β Β (an)m=a(nΓ—m)(a^n )^m=a^{(n\times m)}

We obtain that:Β 
(xβˆ’3)4=Xβˆ’3Γ—4=Xβˆ’12(x^{-3})^4=X^{-3\times 4}=X^{-12}

Great! We got rid of the parentheses! Now we write down the exercise as obtained so far:

xβˆ’3yβˆ’3Γ—xβˆ’12Γ—yβˆ’2yβˆ’4=\frac{x^{-3}}{y^{-3}}\times \frac{x^{-12}\times y^{-2}}{y^{-4}}=

Pay attention that there is a multiplication operation between our two elements!
Therefore, we can join the two elements into one by writing them exactly the same way they are written, just under a fraction line.
We will not forget the multiplication operation between them.
We obtain:
xβˆ’3Γ—xβˆ’12Γ—yβˆ’2yβˆ’3Γ—yβˆ’4=\frac{x^{-3}\times x^{-12}\times y^{-2}}{y^{^{-3}}\times y^{-4}}=
Now we will notice that we are hiding in the exercise the same identical bases between which there is a multiplication operation. Therefore, we can add the relevant powers.
By this rule:Β 
amΓ—an=a(m+n)a^m\times a^n=a^{(m+n)}
We obtain:
xβˆ’15Γ—yβˆ’2yβˆ’7=\frac{x^{-15}\times y^{-2}}{y^{-7}}=

It looks much better now, doesn't it?
Let's continue.
We will notice that in the exercise we received, there is a fraction with the same bases hidden. We can subtract the relevant powers.
According to this rule:Β aman=a(mβˆ’n)\frac {a^m}{a^n} =a^{(m-n)}

We obtain:
xβˆ’15Γ—y5=x^{-15}\times y^5=

Pay attention! We subtract the exponents βˆ’2-2 and βˆ’7-7.
When we subtract a negative from a negative, it becomes positive.

We have reached the final answer, but keep in mind that sometimes you may be asked to show the answer only with positive powers.
We can convert the negative power βˆ’15-15 into a positive power according to this rule:

aβˆ’n=1ana^{-n}=\frac {1}{a^n}

We obtain:
1x15Γ—y5=\frac{1}{x^{15}}\times y^5=
We multiply and obtain:
y5x15\frac{y^5}{x^{15}}


Do you know what the answer is?

Decomposing Natural Numbers into a Product of Powers

We can break down natural numbers and present them as a product of prime factors.
Shall we start with what a natural number is?
A natural number is any positive whole number. There are natural numbers greater than 11 that are divisible without a remainder only by themselves and by 11. These are prime numbers.
For example, let's take the number 22.
22 is divisible by 22 (itself) and also by 11. We cannot find another number in which 22 is divisible without any remainder.
The same is true for the number 33 or 153153, for example. Therefore, these numbers will be called primes.

Example of a non-prime number: 66
66 can be divided by 66, also by 33, by 11, and by 22. It is divisible by more than two numbers and, therefore, is not prime.

What does it mean to decompose a natural number into a product of prime factors?

Let's see this in an example:
Let's take the number 88

88 is a natural number

We can say that:
8=2Γ—2Γ—28=2\times 2\times 2

22 is a prime number. It is divisible by itself and also by 11 only.

And according to the laws of exponents:
8=238=2^3
Basically, we decomposed the 88 into a set of prime numbers. In this case, 22.


How is a natural number decomposed into a product of prime factors?

We will write the number we want to decompose and draw a line next to it.
We will start decomposing it from the smallest initial factor 22.
We ask ourselves, is the number we want to decompose divisible by 22? If so, we will write 22 in front of our number and below the number the result we obtained when we divided it by 22. We will continue asking, what is now the smallest prime number by which our number is divisible without any remainder?
If indeed the number is divisible by a factor without remainder, we will write the prime factor in front of our number and below our number the result we obtained when we divided it by the prime factor.
We will continue decomposing the number until we reach the number 11 in the left column which cannot be decomposed further.
The product of the factors is what we mentioned on the left and is equal to the natural number we decomposed.
Let's see this in an example:

We will take the number 3636
and draw a line next to it:
Think of the smallest prime number: 22.
Question: Is 3636 divisible by 22?
Yes.
Therefore, we will write 22 on the right side of the line and on the left side of the line what remains from the division.

After the division, we will have 1818. We will write 1818 on the left.
Now, let's take the smallest initial factor again and ask, is 1818 divisible by 22?
The answer is yes. After the division, we get 99.
Therefore, we will write 22 again on the right and 99 on the left:
Now, once again we choose the smallest prime number 22 and ask ourselves: Is 99 divisible by 22 without remainder?
The answer is no.
So we will take the next prime number 33 and ask ourselves:
Is 99 divided by 33 without remainder?
The answer is yes.
Therefore, we will add 33 on the right row and on the left, what remains of 99 after dividing by 33, that is 33.
Now we are asked, what is the smallest prime number by which 33 is divided?
The answer is 33.
Therefore, we will add 33 on the right side and 11 on the left side.
When we get 11 on the left side, we have finished the exercise.
If we perform a multiplication operation between all the elements that appear on the right, we obtain the number we decomposed: 3636.

In fact:
36=2Γ—2Γ—3Γ—336=2\times 2\times 3\times 3
According to the rules of exponents, we can write:
36=22Γ—3236=2^2\times 3^2

When will the decomposition of natural numbers help us solve exercises?
Let's see the following exercise:

6781634Γ—926\frac{6^{78}}{16^{34}\times 9^{26}}

We know that according to the rules of exponents we can subtract powers if we have identical bases that are in a fraction.
But in this exercise, the bases are not the same.
How will we convert the bases to be the same?
With the help of prime factor decomposition!


We will take the element 66 and decompose it into factors:


We will take the 1616 and break it down into factors:


We will take the 99 and break it down into factors:


Now, let's use parentheses and put into the exercise above what we obtained:

(2Γ—3)78(2434)Γ—(3226)\frac{(2\times 3)^{78}}{(2^{4}34)\times (3^{2}26)}

Excellent! Now we can use the power rules and solve the exercise easily!
We will obtain:

278Γ—3782136Γ—352=\frac{2^{78}\times 3^{78}}{2^{136}\times 3^{52}}=

We subtract powers and obtain:
2βˆ’58Γ—3262^{-58}\times 3^{26}

Check your understanding

Exponential Equations

What are exponential equations?
Exponential equations are equations in which the unknown appears as an exponent.
Tutorela is here to teach you how to solve exponential equations easily and in different ways.
When you encounter an exponential equation, you will need to arrange it into a certain type and then solve it.
We will start with the first type:


Identical Bases

When
ax=aya^x=a^y
then
x=yx=y

When you encounter an equation where the bases are identical but the exponents are different, you can compare the exponents to solve the equation.
Pay attention! In most equations, you can make the bases identical with the correct factorization!

What's the logic behind this?
This is an equation! Its two sides are equal. Therefore, if the bases are identical, the powers must be equal to each other because they bring the bases to the same result.

Let's see this in an example:
2xβˆ’1=82x2^{x-1}=8^{2x}

We can see that the bases here are not the same.
But, we have already learned that it is possible to decompose a natural number into the product of prime factors.
In fact, we can represent 88 with the help of 22 and thus make the equation have the same bases.
When the bases are identical, we can compare the powers.
First: we will divide the 8 into factors:


8=238=2^3
Now, let's represent what we obtained in the equation and we will get:

2xβˆ’1=(23)2x2^{x-1}=(2^3)^{2x}

According to the rules of exponents we obtain:
2xβˆ’1=26x2^{x-1}=2^{6x}

Excellent! We have reached a state where both bases are equal!
Now we can compare the exponents and solve the equation.
We will obtain:
xβˆ’1=6xx-1=6x

5x=βˆ’15x=-1
x=βˆ’15x=-\frac{1}{5}


Do you think you will be able to solve it?

Tips for Solving in This Way:

First, try to make the bases identical.

  • Remember that when you encounter such bases 2,4,8,16,32,642,4,8,16,32,64 you can represent them in the same base by breaking them down into factors of base 22.

3,9,27,813,9,27,81 can be converted to base 33.
5,25,1255,25,125 can be converted to base 55.

  • Sometimes you will encounter a fraction on one side of the equation. Try to see if you can use this rule:

aβˆ’n=1ana^{-n}=\frac {1}{a^n}
Sometimes, you can move the denominator to the numerator, change the sign of the exponent, and achieve the same bases.
Use a rule that combines powers and roots, and try to convert the root into a power.
anm=anma^{\frac{n}{m}}=\sqrt[m]{a^n}

  • When the number 11 appears on one side of the equation, you can convert it into any base power you want with an exponent of 00.
  • When we fail to reach a state where the two bases are equal on both sides of the equation, we move on to another method.

Solving Exponential Equations Using a Quadratic Equation by Substituting (t) When we encounter exponential equations, we can sometimes use the strategy of transforming them into quadratic equations to find the solution. This method is particularly useful when the exponential equation has a structure that resembles a quadratic, such as having terms that can be considered as squares and square roots in the context of exponents. To apply this technique, we start by identifying a base that appears in all the exponential terms of the equation. Once we have identified this base, we can substitute a variable, often (t), for the base raised to a power. This substitution turns the exponential equation into a quadratic equation in terms of (t). For example, if we have an equation like \(2^{2x} + 2^x - 6 = 0\), we can make the substitution \(t = 2^x\). This transforms our equation into \(t^2 + t - 6 = 0\), which is a quadratic equation. We can then solve for (t) using the quadratic formula or factoring, and afterwards, we solve for (x) by reversing the substitution. It's important to remember that after solving the quadratic equation for (t), we must revert back to our original variable (x) by solving \(2^x = t\). This step may require the use of logarithms to find the value of (x). This method is a powerful tool for solving certain types of exponential equations, and it's a great example of how different areas of mathematics can be used together to solve problems.

When we want to solve exponential equations using a quadratic equation, we need to bring our exercise to a state where there is one element, the same raised to the square, and a free number (without unknowns).
Once we reach this state, we will use a reinforcement player tt "the unknown power".
Thus, we will arrive at an ordinary quadratic equation that we can easily solve.
Let's not forget to place the values of tt that we find to find the values of the unknown we are looking for XX.
Remember the following rules:
(an)m=a(nΓ—m)(a^n )^m=a^{(n\times m)}
amΓ—an=a(m+n)a^m\times a^n=a^{(m+n)}
and let's take the following example:
42x+4x+1βˆ’12=04^{2x}+4^{x+1}-12=0
To bring the equation to a state where there is one element, the same squared, and a free number, we will have to use power rules.
According to the power rules we can say that:
42x=(4x)24^{2x}=(4^x)^2
Basically, "we convert the rule" to "isolate" 4x4^x and show that it is squared.

We will also take 4x+14^{x+1}
According to the power rules we can say that:
4x+1=4Γ—4x4^{x+1}=4\times 4^x
Now, let's rewrite the same equation with the data we received and we get:
(4x)2+4Γ—4xβˆ’12=0(4^x)^2+4\times 4^x-12=0
Wonderful! We have reached a state where there is an equation with one element. 4x4^x The same element squared and a free number.
What do we do now?
Now we will bring in a reinforcement player. To keep it simple, we will place the t instead of the element with the unknown power incΓ³gnita.
Let's say that: t=4xt=4^x
Every time we use 4x4^x We will replace it with tt
Now, our equation will be simpler and will look like this:

t2+4Γ—tβˆ’12=0t^2+4\times t-12=0
Indeed, we have reached a simple quadratic equation
Solve it and get that:
t2+4tβˆ’12=0t^2+4t-12=0
t=βˆ’6,t=2t=-6, t= 2
Pay attention! This is not our final answer!
We found tt and not the unknown we are looking for xx
To find the value of x in the equation t=4xt=4^x Once t=βˆ’6t=-6 And another time t=2t=2 We get:

When t=βˆ’6t=-6
4x=βˆ’64^x=-6
The equation has no real solution, as the powers of a positive number cannot yield a negative number
When t=2t=2
4x=24^x=2
x=0.5x=0.5

Therefore, the result is: x=0.5x=0.5


Test your knowledge

Scientific Notation

In certain scientific subjects like biology and chemistry, for example, there are very large or very small numbers.
We can express them in a readable and convenient way, through scientific notation.
In scientific notation, the number is presented as a product of a number that is between 11 and 1010 multiplied by 1010 to some power.
Generally, the expression will look like this:

mΓ—10em\times 10^e

mm will be a number between 00 and 11.
If ee is a positive integer, the expression will be a number greater than 11.
If ee is a negative integer, the expression will be a number less than 11.


If you are interested in this article, you might also be interested in the following articles:

  • The Rules of Exponentiation
  • Division of Powers with the Same Base
  • Power of a Multiplication
  • Power of a Quotient
  • Power of a Power
  • Power with Zero Exponent
  • Powers with a Negative Integer Exponent
  • Taking Advantage of All the Properties of Powers or the Laws of Exponents
  • Exponentiation of Whole Numbers

On the Tutorela blog, you will find a variety of articles about mathematics.


Exercises on Powers

Exercise 1:

Problem Statement

82β‹…83β‹…85= 8^2\cdot8^3\cdot8^5=

Solution

All the bases are equal and therefore the exponents can be added.

82β‹…83β‹…85=810 8^2\cdot8^3\cdot8^5=8^{10}

Answer

810 8^{10}


Exercise 2:

Problem Statement

Simplify the expression:

a3β‹…a2β‹…b4β‹…b5= a^3\cdot a^2\cdot b^4\cdot b^5=

Solution

In this case we have 2 different bases a and b. We will add the exponents of each base separately.

a3β‹…a2β‹…b4β‹…b5=a5β‹…b9 a^3\cdot a^2\cdot b^4\cdot b^5=a^5\cdot b^9

Answer

a5β‹…b9 a^5\cdot b^9


Exercise 3:

Problem Statement

3xβ‹…2xβ‹…32x= 3^x\cdot2^x\cdot3^{2x}=

Solution

In this case we have 2 different bases, so we will add what can be added, that is, the exponents of 3 3

3xβ‹…2xβ‹…32x=2xβ‹…33x 3^x\cdot2^x\cdot3^{2x}=2^x\cdot3^{3x}

Answer

33xβ‹…2x 3^{3x}\cdot2^x


Exercise 4:

Problem Statement

22x+1β‹…25β‹…23x= 2^{2x+1}\cdot2^5\cdot2^{3x}=

Solution

Since the bases are equal, the exponents can be added:

2x+1+5+3x=5x+6 2x+1+5+3x=5x+6

Answer

25x+6 2^{5x+6}


Exercise 5:

Problem Statement

Solve the exercise by factoring out the common factor

6x6βˆ’9x4=0 6x^6-9x^4=0

Solution

First we take out the smallest power

6x6βˆ’9x4= 6x^6-9x^4=

6x4(x2βˆ’1.5)=0 6x^4\left(x^2-1.5\right)=0

If possible we reduce the numbers by a common denominator

Finally we will compare both sections with: 0 0

6x4=0 6x^4=0

We divide by: 6x3 6x^3

x=0 x=0

x2βˆ’1.5=0 x^2-1.5=0

x2=1.5 x^2=1.5

x=Β±32 x=\pm\sqrt{\frac{3}{2}}

Answer

x=0,x=Β±32 x=0,x=\pm\sqrt{\frac{3}{2}}


Exercise 6:

Problem Statement

((8by)3)y+(3x)a= ((8by)^3)^y+(3^x)^a=

Solution

(8by)3β‹…y+3xβ‹…a \left(8by\right)^{3\cdot y}+3^{x\cdot a}

First we use the law

(am)n=amβ‹…n \left(a^m\right)^n=a^{m\cdot n}

After that we will expand the parentheses according to the law.

(abc)x=axβ‹…bxβ‹…cx \left(abc\right)^x=a^x\cdot b^x\cdot c^x

83yβ‹…b3yβ‹…y3y+3xa 8^{3y}\cdot b^{3y}\cdot y^{3y}+3^{xa}

Answer

83yβ‹…b3yβ‹…y3y+3xa 8^{3y}\cdot b^{3y}\cdot y^{3y}+3^{xa}


Do you know what the answer is?

Review Questions

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What is a power and what are its components?

A power consists of two components: a base and an exponent.

The base is the number that will be multiplied, while the exponent is the number that indicates how many times the base will be multiplied.


What is the difference between a prime number and a composite number?

A prime number has only two divisors, one and itself, while a composite number has more than two divisors.

Examples of prime numbers:

2,3,5,7,11,13,17,19,23,29…2, 3, 5, 7, 11, 13, 17, 19, 23,29…

These numbers can only be divided by 1 1 and by themselves.

Examples of composite numbers:

As already mentioned, these are the ones that have more than two divisors, therefore they are the numbers that do not appear in the previous example

4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26…4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26…


What is a power and some examples?

A power is one that is composed of a base and an exponent as shown:

an a^n

Where a a is called the base and n n is the exponent. The base is the number that is multiplied and the exponent indicates how many times it is multiplied.

Let's see some examples:

35=3Γ—3Γ—3Γ—3Γ—3=243 3^5=3\times3\times3\times3\times3=243

57=5Γ—5Γ—5Γ—5Γ—5Γ—5Γ—5=78125 5^7=5\times5\times5\times5\times5\times5\times5=78125

93=9Γ—9Γ—9=729 9^3=9\times9\times9=729

x4=xΓ—xΓ—xΓ—x x^4=x\times x\times x\times x

z2=zΓ—z z^2=z\times z

We can observe that powers are valid for algebraic expressions.


What are the types of powers in mathematics?

There are 7 types of power rules, which are called the laws of exponents, which are as follows:

  • Multiplication of powers with the same base

amΓ—an=a(m+n) a^m\times a^n=a^{\left(m+n\right)}

When we have the multiplication of powers with the same base, the result will be the same base and the exponents are added.

  • Quotient of powers with the same base

aman=a(mβˆ’n) \frac{a^m}{a^n}=a^{\left(m-n\right)}

Here the exponents are subtracted, the result will be the same base and we subtract the exponent of the numerator minus that of the denominator.

  • Multiplication of different bases with the same exponent

anΓ—bn=(aΓ—b)n a^n\times b^n=\left(a\times b\right)^n

In this law, we can observe that the exponent can be distributed to each element of the multiplication.

  • Quotient of different bases with the same exponent.

anbn=(ab)n \frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n

For this law, the same happens as in the previous one, the exponent is distributed to each of the elements of the fraction.

  • Power of a power

(xm)n=xmΓ—n \left(x^m\right)^n=x^{m\times n}

When we have a power raised to another power, the exponents must be multiplied.

  • Zero power. Any number other than zero, raised to a zero exponent, will always result in 1 1 .

a0=1 a^0=1

  • Negative power.

aβˆ’n=1an a^{-n}=\frac{1}{a^n}

We can observe that we have a negative exponent, so we can write it as its reciprocal but with a positive exponent.


How can we use some of these laws of exponents in examples?

To see how to apply certain laws, let's look at some examples:

Example 1.

34Γ—3Γ—32= 3^4\times3\times3^2=

Here we can observe that we have a multiplication of powers with the same base, so we will use the first law, where the result will be the same base and we add the exponents:

4+1+2=7 4+1+2=7

Therefore the result is:

37 3^7

Example 2.

(2Γ—3)5(22Γ—3)4= \frac{\left(2\times3\right)^5}{\left(2^2\times3\right)^4}=

To solve this exercise we must apply not just one law, but a combination of the laws we have already studied, in the numerator and denominator we can use the third law of exponents, which is the multiplication of powers with different bases but the same exponent, we will do it separately.

(2Γ—3)5=25Γ—35 \left(2\times3\right)^5=2^5\times3^5

(22Γ—3)4=(22)4Γ—34 \left(2^2\times3\right)^4=\left(2^2\right)^4\times3^4

In the denominator we can see that there is a power to another power, so we can simplify it by multiplying the exponents, resulting in the following:

25Γ—3528Γ—34= \frac{2^5\times3^5}{2^8\times3^4}=

Now we apply the second law, of the quotient of the same base, just subtracting the exponents of the equal bases:

25Γ—3528Γ—34=2(5βˆ’8)Γ—3(5βˆ’4)=2βˆ’3Γ—31 \frac{2^5\times3^5}{2^8\times3^4}=2^{\left(5-8\right)}\times3^{\left(5-4\right)}=2^{-3}\times3^{1}

Finally, we use the seventh law, since we have a negative exponent.

2βˆ’3Γ—3=123Γ—3=38 2^{-3}\times3=\frac{1}{2^3}\times3=\frac{3}{8}

Therefore the result is

38 \frac{3}{8}


What is the difference between power and multiplication?

Power is a way of writing in abbreviated form the multiplication of a term by itself several times. The figure that is multiplied by itself is called the base, while the number of times it is multiplied by itself is called the exponent.


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