Zero Exponent Rule

🏆Practice powers with zero exponent

When we see a number that is not 0 0 raised to zero, the result will be 1 1 .
Property formula:

a0=1a^0=1
This property is also concerning algebraic expressions.

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Test yourself on powers with zero exponent!

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\( 5^0= \)

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Examples with explanations for zero exponents

Examples:
50=15^0=1
(2)0=1(-2)^0=1
x0=1x^0=1


The property of the zero exponent is simple and easy since, regardless of the number or unknown (with the exception of 0 0 ), raising it to 0 0 will give 1 1 .

However, like anything in life, when we understand its logic it becomes easier to remember.

So, let's talk a bit about the logic behind this priority.

We promise you it will be worth it.

Why does every number raised to 0 0 is equal to 1 1 ?

This property is quite intuitive for the simple fact that, if we observe that any base is increasing as we raise it to different powers, if we go back we will see that the smaller the exponent the base will decrease relatively.


For example, if we take the base 2 2 :

21,22,23,24 2^1,2^2,2^3,2^4

The value doubles as the exponent grows by one:

2,4,8,16.... 2,4,8,16....

Now we will ask ourselves what would happen if we decrease the exponent of 21 2^1 by one unit and start from 20 2^0 ?

The value decreases by half, right?

From 2 2 it would become 1 1 .

So it makes a lot of sense to us to think that 20=1 2^0=1


Another way we can see the logic behind this property is by looking at the following expression

When a0 a≠0 :

anan \frac{a^n}{a^n}

We know that any number divided by itself is equal to 1 1 therefore, it is true that:

anan=1 \frac{a^n}{a^n}=1

Furthermore, we have learned that if we have a fraction with equal bases we can subtract the exponents.

That is, it will give us:

anan=ann \frac{a^n}{a^n}=a^{n-n}

Let's see what we got:ann a^{n-n}

Let's observe the power, every number subtracted by itself gives 0 0 . Therefore, this gives us:

anan=1 \frac{a^n}{a^n}=1

Note that we have previously remembered that:

anan=1 \frac{a^n}{a^n}=1

and that according to the transitive relation gives us:

anan=1=a0 \frac{a^n}{a^n}=1=a^0

Let's see examples of the use of this property as part of the exercise:

X890×03×32= \frac{X^{890\times 0}}{3}\times \frac{3}{2}=

Let's look carefully at the following exercise. We will notice that the first step we have to do is to multiply the exponents that are separated by multiply operations.

We will obtain:

X03×32= \frac{X^0}{3}\times \frac{3}{2}=

Let us remember that the zero exponent rule tells us that every term raised to 0 0 (with the exception of 0 0 ) is equal to 1 1 , therefore, also in the case of dealing with an unknown, as long as X X does not equal 0 0 , the result will be 1 1 . We will obtain:

13×32= \frac{1}{3}\times \frac{3}{2}=

Now we will multiply the fractions normally to have only one fraction:

36=12 \frac{3}{6}=\frac{1}{2}

Let's look at another example that is a bit misleading:

04×50×2210= \frac{0^{-4}\times 5^0\times 2^2}{1^0}=

If we pay attention we will not be fooled at all and we will solve the exercise correctly!

Let's start with the first term, 0 0 raised to any power (outside of 0 0 ) will give us 0 0 , therefore, the first term equals 0 0 .

Stop now! You can continue looking at the whole exercise.... but, when we already know that we have a 0 0 in the numerator that is multiplied by all the other terms in the numerator, we understand that the whole expression is equivalent to 0 0 .

It does not matter at all that the denominator equals 1 1 , as we deduce according to the property, or that: 50 5^0 also equals 1 1 .

so:

04×50×2210= \frac{0^{-4}\times 5^0\times 2^2}{1^0}=

Now let's look at an example in which you will have to clear the X X and you will have to do it using this property:

47668.786X=1 47668.786^X=1

This equation probably seems difficult and extremely complicated, but, if we remember that every number with exponent 0 0 is equivalent to 1 1 , we will deduce that the unknown X X that we are looking for cannot be anything other than 0 0 .

So:

X=0 X=0

Useful information:

If we take 0 0 and raise it to the power of 0 0 it will give us an undefined expression.


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Exercises with zero exponents

Exercise 1

Solve the following exercise:

50= 5^0=

Solution:

Any number with zero exponent is equal to. 1 1

Answer:

50=1 5^0= 1


Exercise 2

Solve the following exercise:

X0=? X^0=?

Solution:

Also when we refer to unknowns, any number with exponent 0 0 is still worth 1 1 .

Answer:

X0=1 X^0=1


Do you know what the answer is?

Exercise 3

Simplify the following expression:

4223 \frac{4\cdot2}{2^3}

Solution:

We write 44 as 222^2 and apply the power product rule for the same base in the numerator, thus we obtain:

22223=2323\frac{2^2\cdot2}{2^3}=\frac{2^3}{2^3}

Finally, we apply the quotient rule and the zero exponent rule to get the result.

233=20=12^{3-3}=2^0=1

Answer:

4223=1 \frac{4\cdot2}{2^3} =1


Exercise 4

Calculate the value of:

3721+35+24 3^{7-2-1+3-5+2-4}

Solution:

We group the positive and negative numbers found in the exponent, add the numbers with the same sign and obtain:

37+3+22154=31212=30=13^{7+3+2-2-1-5-4} = 3^{12-12} =3^{0}=1

Let's note that the result is 1 1 , because we obtained a power of zero exponent.

Answer:

37+3+22154=1 3^{7+3+2-2-1-5-4} =1


Check your understanding

Exercise 5

Find the value of X X :

5x+225=0 5^{x+2}-25=0

Solution:

We add 2525 to both sides and obtain:

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

We write 25 25 as 52 5^2, and divide both sides of the equality by that number. The result is:

5x+252=5252 \frac{5^{x+2}}{5^2}=\frac{5^2}{5^2}

5x+22=1 5^{x+2-2}=1

5x=1 5^{x}=1

From the last equation , we can deduce that x=0 x=0 , because a non-zero number raised to zero is equal to 1 1.

Answer:

x=0x=0


Review Questions

What happens when the exponent is zero?

If a non-zero number is raised to zero, the result is 1 1 . To check this result we can apply on the one hand the law of exponents for quotients of the same base and on the other hand use the fact that if in a quotient the numerator and denominator are equal the result is 1 1 .


What happens when the exponent is less 1 1 ?

When a non-zero number is raised to minus 1 1 , then we get the multiplicative inverse of that number.

a1=1a a^{-1}=\frac{1}{a}


Why isn't 0 0 raised to 0 0 also always 1 1 ?

In general there is no consensus on the result of 0 0 raised to 0 0 , in some branches of mathematics this expression is worth 1 1 , while in others the result of the expression is not defined.


Do you think you will be able to solve it?

examples with solutions for powers with zero exponent

Exercise #1

50= 5^0=

Video Solution

Step-by-Step Solution

We use the power property:

X0=1 X^0=1 We apply it to the problem:

50=1 5^0=1 Therefore, the correct answer is C.

Answer

1 1

Exercise #2

1120=? 112^0=\text{?}

Video Solution

Step-by-Step Solution

We use the zero exponent rule.

X0=1 X^0=1 We obtain

1120=1 112^0=1 Therefore, the correct answer is option C.

Answer

1

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

(7125)0=? (\frac{7}{125})^0=\text{?}

Video Solution

Step-by-Step Solution

We use the zero exponent rule.

X0=1 X^0=1 We obtain:

(7125)0=1 \big( \frac{7}{125}\big)^0=1 Therefore, the correct answer is option B.

Answer

1

Exercise #5

(74)?=1 (\frac{7}{4})^?=1

Video Solution

Step-by-Step Solution

We use the fact that raising any number (except zero) to the power of zero will yield the result 1:

X0=1 X^0=1 Therefore, it is clear that:

(74)0=1 (\frac{7}{4})^0=1 Therefore, the correct answer is option C.

Answer

0

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