Fraction power - Examples, Exercises and Solutions

Power of a Quotient

When we encounter an expression with a quotient (or division) inside parentheses and the entire expression is raised to a certain exponent, we can take the exponent and apply it to each of the terms in the expression.
Let's not forget to maintain the fraction bar between the terms.
Formula of the property:
(ab)n=anbn(\frac {a}{b})^n=\frac {a^n}{b^n}
This property is also relevant to algebraic expressions.

Suggested Topics to Practice in Advance

  1. Power of a Power
  2. Exponent of a Multiplication

Practice Fraction power

Exercise #1

(4274)2= (\frac{4^2}{7^4})^2=

Video Solution

Step-by-Step Solution

(4274)2=42×274×2=4478 (\frac{4^2}{7^4})^2=\frac{4^{2\times2}}{7^{4\times2}}=\frac{4^4}{7^8}

Answer

4478 \frac{4^4}{7^8}

Exercise #2

(26)3= (\frac{2}{6})^3=

Video Solution

Step-by-Step Solution

We use the formula:

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

(26)3=(22×3)3 (\frac{2}{6})^3=(\frac{2}{2\times3})^3

We simplify:

(13)3=1333 (\frac{1}{3})^3=\frac{1^3}{3^3}

1×1×13×3×3=127 \frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}

Answer

127 \frac{1}{27}

Exercise #3

54(15)4=? 5^4\cdot(\frac{1}{5})^4=\text{?}

Video Solution

Step-by-Step Solution

This problem can be solved using the properties of powers for a negative power, power over a power, and the property of powers for the product between terms with identical bases, which is the natural way of solving,

But here we prefer to solve it in another way that is a bit faster:

To this end, the power by power law is applied to the parentheses in which the terms are multiplied, but in the opposite direction:

xnyn=(xy)n x^n\cdot y^n=(x\cdot y)^n Since in the expression in the problem there is a multiplication between two terms with identical powers, this law can be used in its opposite sense, so we will apply this property to the problem:

54(15)4=(515)4 5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4 Since the multiplication in the given problem is between terms with the same power, we could apply this law in the opposite direction and write the expression as the multiplication of the bases of the terms in parentheses to which the same power is applied.

We will continue and simplify the expression in parentheses, we will do it quickly if we notice that in parentheses there is a multiplication between two opposite numbers, then their product will give the result: 1, we will apply this understanding to the expression we arrived at in the last step:

(515)4=14=1 \big(5\cdot\frac{1}{5}\big)^4 = 1^4=1 When in the first step we apply the previous understanding, and then use the fact that raising the number 1 to any power will always give the result: 1, which means that:

1x=1 1^x=1 Summarizing the steps to solve the problem, we get that:

54(15)4=(515)4=1 5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4 =1 Therefore, the correct answer is option b.

Answer

1

Exercise #4

7483(17)4=? 7^4\cdot8^3\cdot(\frac{1}{7})^4=\text{?}

Video Solution

Step-by-Step Solution

We use the formula:

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

We decompose the fraction in parentheses:

(17)4=1474 (\frac{1}{7})^4=\frac{1^4}{7^4}

We obtain:

74×83×1474 7^4\times8^3\times\frac{1^4}{7^4}

We simplify the powers: 74 7^4

We obtain:

83×14 8^3\times1^4

Remember that the number 1 in any power is equal to 1, so we obtain:

83×1=83 8^3\times1=8^3

Answer

83 8^3

Exercise #5

(23)4=? (\frac{2}{3})^{-4}=\text{?}

Video Solution

Step-by-Step Solution

We use the formula:

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

Therefore, we obtain:

(32)4 (\frac{3}{2})^4

We use the formula:

(ba)n=bnan (\frac{b}{a})^n=\frac{b^n}{a^n}

Therefore, we obtain:

3424=3×3×3×32×2×2×2=8116 \frac{3^4}{2^4}=\frac{3\times3\times3\times3}{2\times2\times2\times2}=\frac{81}{16}

Answer

8116 \frac{81}{16}

Exercise #1

9?(12)4=163 9^?(\frac{1}{2})^{-4}=\frac{16}{3}

Video Solution

Step-by-Step Solution

We address the problem:

9?(12)4=163 9^?(\frac{1}{2})^{-4}=\frac{16}{3} as an equation for all (and of course it is indeed an equation),

Therefore, we replace the problem sign in the unknown x and solve it:

9x(12)4=163 9^x(\frac{1}{2})^{-4}=\frac{16}{3}

Now we briefly discuss the solution technique:

Quite generally, the goal when solving exponential equations is to reach a situation where there is a term on each side of the equation so that both sides have the same base, in such a situation we can unequivocally state that the power exponents on both sides of the equation are equal, and solve a simple equation for the unknown,

Mathematically, we will perform a mathematical manipulation (according to the laws of course) on both sides of the equation (or development of one side with the help of power properties and algebra) and will arrive at the following situation:

bm(x)=bn(x) b^{m(x)}=b^{n(x)} whenm(x),n(x) m(x),\hspace{4pt}n(x) Algebraic expressions (actually functions of the unknownx x ) that can also exclude the unknowns (x x ) that we try to find in the problem, which is the solution to the equation,

It is then stated that:

m(x)=n(x) m(x)=n(x) and we solve the simple equation we get,

We solve the equation in the given problem again:

9x(12)4=163 9^x(\frac{1}{2})^{-4}=\frac{16}{3} In solving this equation, various power properties are used:

a. Power property with negative exponent:

an=1an a^{-n}=\frac{1}{a^n} b. Power property for a power of an exponent raised to another exponent:

(am)n=amn (a^m)^n=a^{m\cdot n}

First we will arrive at a simple presentation of the terms of the equation, that is, "eliminate" fractions and roots (if there are any in the problem, there are none here)

To do this, we start by treating the fraction on the right side of the equation, this will be done using the power property of a negative exponent specified in A above and represent this fraction (in parentheses) as a term with a negative exponent:

9x(12)4=1639x(21)4=1639x2(1)(4)=1639x24=163 9^x(\frac{1}{2})^{-4}=\frac{16}{3} \\ 9^x(2^{-1})^{-4}=\frac{16}{3}\\ 9^x2^{(-1)\cdot(-4)}=\frac{16}{3}\\ 9^x2^{4}=\frac{16}{3}\\ When we perform the development on the left side of the equation as described above, and in the last step simplify the expression in the power exponent on the left side of the equation,

Later we would like to be able to obtain an identical base on both sides of the equation, the best way to achieve this is by decomposing all and every one of the numbers in the problem into prime factors (using powers as well), here in the problem we note that the numbers exist:

16,9,3,2 16,\hspace{4pt}9,\hspace{4pt}3,\hspace{4pt}2 The numbers: 2, 3 are prime, so we will not touch them, we will note that the number 16 is a power of the number 2 and that the number 9 is a power of the number 3, that is:

16=249=32 16=2^4\\ 9=3^2 This is the presentation (decomposition) of the numbers 16 and 9 with the help of their prime factors, so we will return to the equation we obtained in the previous step and replace these numbers in the decomposition of their prime factors:

9x24=163(32)x24=243 9^x2^{4}=\frac{16}{3}\\ (3^2)^x2^{4}=\frac{2^4}{3}\\ Now we will notice that we can get rid of the term.24 2^4 By dividing both sides of the equation by it, we will also notice that this term does not depend on the unknown and is different from zero and therefore there is no limitation that says it is forbidden to divide it, we will do so:

(32)x24=243/:24(32)x̸24̸24≠243̸24(32)x=13 (3^2)^x2^{4}=\frac{2^4}{3} \hspace{8pt}\text{/:}2^{4}\\ \frac{(3^2)^x\cdot\not{2^4}}{\not{2^4}}=\frac{\not{2^4}}{3\cdot\not{2^4}} \\ (3^2)^x=\frac{1}{3} When in the first step we divide both sides of the equation by the term we wanted to get rid of and then simplify the fractions obtained on both sides of the equation,

Now we go back to remember the power laws that we have already used and that were mentioned before:

a. Power property with negative exponent:

an=1an a^{-n}=\frac{1}{a^n} b. Power property for a power of an exponent raised to another exponent:

(am)n=amn (a^m)^n=a^{m\cdot n} In the next step, we will apply the power raised to another power law specified in B above on the left side, to get rid of the parentheses, and in the next step we deal with the right side with the goal of undoing the fraction, for this purpose, we will use the power property with a negative exponent specified in A above, we will perform a step by development line:

(32)x=1332x=1332x=31 (3^2)^x=\frac{1}{3} \\ 3^{2x}=\frac{1}{3} \\ 3^{2x}=3^{-1} We have reached our goal, we obtained an equation in which both sides have terms with the same base, therefore we can affirm that the power exponents of the terms on both sides are equal, and to solve the resulting equation for the unknown, we do the following:

32x=312x=1 3^{2x}=3^{-1} \\ \downarrow\\ 2x=-1 We continue and solve the resulting equation, this we will do by isolating the unknown on the left side, we will achieve this by dividing both sides of the equation by its coefficient:

2x=1/:2x=12 2x=-1 \hspace{8pt}\text{/:}2 \\ \bm{x=-\frac{1}{2} } We have thus solved the given equation, we briefly summarize the solution steps:9x(12)4=1639x24=163(32)x24=243/:24(32)x=1332x=312x=1/:2x=12 9^x(\frac{1}{2})^{-4}=\frac{16}{3} \\ 9^x2^{4}=\frac{16}{3}\\ (3^2)^x2^{4}=\frac{2^4}{3}\hspace{8pt}\text{/:}2^{4}\\ (3^2)^x=\frac{1}{3} \\ 3^{2x}=3^{-1} \\ \downarrow\\ 2x=-1\hspace{8pt}\text{/:}2 \\ \bm{x=-\frac{1}{2} } Therefore, the correct answer is option c.

Answer

12 -\frac{1}{2}

Exercise #2

((15)2)?:5=125 ((\frac{1}{5})^2)^?:5=125

Video Solution

Step-by-Step Solution

We address the problem:

((15)2)?:5=125 \big( \big(\frac{1}{5} \big)^2 \big)^?:5=125 as an equation for all (and of course it is indeed an equation),

Therefore, we replace the problem sign in the unknown x and solve it:

((15)2)x:5=125 \big( \big(\frac{1}{5} \big)^2 \big)^x:5=125 Later on we will remember that dividing by a certain number is multiplying by its inverse, so we will rewrite the given equation bearing this in mind:

((15)2)x15=125 \big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125

Now we briefly discuss the solution technique:

In quite a general way, the goal when solving exponential equations is to achieve a situation where there is a term on each of the two sides of the equation so that both sides have the same base, in such a situation we can unequivocally state that the power exponents on both sides of the equation are equal, and solve a simple equation for the unknown,

Mathematically, we will perform a mathematical manipulation (according to the laws of course) on both sides of the equation (or development of one of the sides with the help of power properties and algebra) and arrive at the following situation:

bm(x)=bn(x) b^{m(x)}=b^{n(x)} when m(x),n(x) m(x),\hspace{4pt}n(x) Algebraic expressions (actually functions of the unknown x x ) that can also exclude the unknowns (x x ) that we are trying to find in the problem, which is the solution to the equation,

It is then stated that:

m(x)=n(x) m(x)=n(x) and we solve the simple equation we obtain,

We go back to solving the equation in the given problem:

((15)2)x15=125 \big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 In solving this equation, various power properties are used:

a. Power property with negative exponent:

an=1an a^{-n}=\frac{1}{a^n} b. Power property for a power of an exponent raised to another exponent:

(am)n=amn (a^m)^n=a^{m\cdot n}

First we will arrive at a simple presentation of the terms of the equation, that is, "eliminate" fractions and roots (if there are any in the problem, there are none here)

To do this, we will start by dealing with the fraction on the left side of the equation:

15 \frac{1}{5} That is, both the fraction inside the parenthesis and the fraction outside the parenthesis, this is done with the help of the power property with negative exponent specified in A above and we represent this fraction as a term with negative power and in the next step we will apply the power property for a power of an exponent raised to another exponent specified in B above and we will get rid of the parentheses, another step starting from the inner parenthesis to the outer, we will do this, step by step below:

((15)2)x15=125((51)2)x51=125(5(1)2)x51=1255(1)2x51=12552x51=125 \big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ (5^{(-1)\cdot 2} )^x\cdot 5^{-1}=125 \\ 5^{(-1)\cdot 2\cdot x} \cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\ When we carry out the development of the left side of the equation as described above, we initially apply the power property with negative exponent mentioned above in A and in the following steps we apply the power property for a power of an exponent raised to another exponent mentioned above in B and we got rid of the parentheses: starting from the inner parenthesis to the outer, in the last step we simplify the expression in the power exponent on the left side of the equation,

c. Later we remember the power property for multiplying terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n} And we will apply this law to the left side of the equation we obtained in the last step, this is to have a term on this side, we will do this:

52x51=12552x+(1)=12552x1=125 5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x+(-1)}=125 \\ 5^{-2x-1}=125 \\ When in the first step we apply the aforementioned power law to the product between members with identical bases mentioned above in C and in the following steps we simplify the expression in the power exponent on the left side,

Next, we would like to obtain the same base on both sides of the equation, the best way to achieve this is by decomposing each and every one of the numbers in the problem into prime factors (using powers as well), you will notice that the number 125 is a power of the number 5, that is:

125=53 125=5^3 This is the presentation (factorization) of the number 125 using its prime factor, which is the number 5, so we go back to the equation we received in the previous step and replace this number with its decomposition into prime factors:

52x1=12552x1=53 5^{-2x-1}=125 \\ 5^{-2x-1}=5^3 \\ We have reached our goal, we have received an equation in which both sides have terms with the same base, therefore we can state that the power exponents of the terms on both sides are equal, and to solve the resulting equation for the unknown, we will do this:

52x1=532x1=3 5^{-2x-1}=5^3 \\ \\ \downarrow\\ -2x-1=3 We will continue and solve the resulting equation, we will do this by isolating the unknown on the left side, we will achieve this in the usual way, moving the sections and dividing the final equation by the unknown's coefficient:

2x1=32x=3+12x=4/:(2)̸2x̸2=42x=42x=2 -2x-1=3 \\ -2x=3+1\\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \frac{\not{-2}x}{\not{-2}}=\frac{4}{-2}\\ x=-\frac{4}{2}\\ \bm{x=-2 } When in the first step we simplify the equation by moving the sides, remembering that when a term is moved its sign changes, then we complete the isolation by nullifying dividing both sides of the equation by its coefficient, in the last steps, we simplify the expression obtained by reducing the fractions,

We have thus solved the given equation, we briefly summarize the solution steps:

((15)2)x15=125((51)2)x51=12552x51=12552x1=532x1=32x=4/:(2)x=2 \big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x-1}=5^3 \\ \downarrow\\ -2x-1=3 \\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \bm{x=-2 } Therefore, the correct answer is option a.

Answer

2 -2

Exercise #3

What is the result of the following power?

(23)3 (\frac{2}{3})^3

Video Solution

Answer

827 \frac{8}{27}

Exercise #4

454614=? 4^5-4^6\cdot\frac{1}{4}=\text{?}

Video Solution

Answer

0

Exercise #5

(132)0(213)2(132)5=? (\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}=\text{?}

Video Solution

Answer

(213)3 (\frac{2}{13})^3

Exercise #1

3004(1300)4=? 300^{-4}\cdot(\frac{1}{300})^{-4}=?

Video Solution

Answer

1

Exercise #2

(78)2=? (\frac{7}{8})^{-2}=\text{?}

Video Solution

Answer

11549 1\frac{15}{49}

Exercise #3

(axb)z=? (\frac{ax}{b})^{-z}=\text{?}

Video Solution

Answer

bzazxz b^za^{-z}x^{-z}

Exercise #4

(37)9=? (\frac{3}{7})^{-9}=\text{?}

Video Solution

Answer

7939 \frac{7^9}{3^9}

Exercise #5

Which value is greater?

Video Solution

Answer

(x3)5 (x^3)^5

Topics learned in later sections

  1. Multiplying Exponents with the Same Base
  2. Division of Exponents with the Same Base
  3. Rules of Exponentiation
  4. Combining Powers and Roots