Zero Exponent Rule - Examples, Exercises and Solutions

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.

Suggested Topics to Practice in Advance

  1. Exponents - Special Cases
  2. Exponents of Negative Numbers

Practice Zero Exponent Rule

Examples with solutions for Zero Exponent Rule

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

(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 #4

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

Video Solution

Step-by-Step Solution

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

X0=1 X^0=1 It is thus clear that:

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

Answer

0

Exercise #5

(3005327)0=? (300\cdot\frac{5}{3}\cdot\frac{2}{7})^0=\text{?}

Video Solution

Step-by-Step Solution

Due to the fact that raising any number (except zero) to the power of zero will give the result 1:

X0=1 X^0=1 Let's examine the expression of the problem:

(3005327)0 (300\cdot\frac{5}{3}\cdot\frac{2}{7})^0 The expression inside of the parentheses is clearly not 0 (it can be calculated numerically and verified)

Therefore, the result of raising to the power of zero will give the result 1, that is:

(3005327)0=1 (300\cdot\frac{5}{3}\cdot\frac{2}{7})^0 =1 Therefore, the correct answer is option A.

Answer

1

Exercise #6

7x7x=? 7^x\cdot7^{-x}=\text{?}

Video Solution

Step-by-Step Solution

We use the law of exponents to multiply terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n} We apply the law to given the problem:

7x7x=7x+(x)=7xx=70 7^x\cdot7^{-x}=7^{x+(-x)}=7^{x-x}=7^0 In the first stage we apply the above power rule and in the following stages we simplify the expression obtained in the exponent,

Subsequently, we use the zero power rule:

X0=1 X^0=1 We obtain:

70=1 7^0=1 Lastly we summarize the solution to the problem.

7x7x=7xx=70=1 7^x\cdot7^{-x}=7^{x-x}=7^0 =1 Therefore, the correct answer is option B.

Answer

1 1

Exercise #7

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 inside of the 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, thus we obtain:

83×1=83 8^3\times1=8^3

Answer

83 8^3

Exercise #8

9380=? \frac{9\cdot3}{8^0}=\text{?}

Video Solution

Step-by-Step Solution

We use the formula:

a0=1 a^0=1

9×380=9×31=9×3 \frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3

We know that:

9=32 9=3^2

Therefore, we obtain:

32×3=32×31 3^2\times3=3^2\times3^1

We use the formula:

am×an=am+n a^m\times a^n=a^{m+n}

32×31=32+1=33 3^2\times3^1=3^{2+1}=3^3

Answer

33 3^3

Exercise #9

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

Video Solution

Step-by-Step Solution

This problem can be solved using the Law of exponents power rules for a negative power, power over a power, as well as the power rule for the product between terms with identical bases.

However we prefer to solve it in a quicker way:

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.

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 can 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 continue and simplify the expression inside of the parentheses. We can do it quickly if inside the parentheses there is a multiplication between two opposite numbers, then their product will give the result: 1, All of the above is applied to the problem leading us to the last step:

(515)4=14=1 \big(5\cdot\frac{1}{5}\big)^4 = 1^4=1 We remember 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 obtain the following:

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 #10

53505255= 5^{-3}\cdot5^0\cdot5^2\cdot5^5=

Video Solution

Step-by-Step Solution

We use the power property to multiply terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n} Keep in mind that this property is also valid for several terms in the multiplication and not just for two, for example for the multiplication of three terms with the same base we obtain:

amanak=am+nak=am+n+k a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k} When we use the mentioned power property twice, we could also perform the same calculation for four terms of the multiplication of five, etc.,

Let's return to the problem:

Keep in mind that all the terms of the multiplication have the same base, so we will use the previous property:

53505255=53+0+2+5=54 5^{-3}\cdot5^0\cdot5^2\cdot5^5=5^{-3+0+2+5}=5^4 Therefore, the correct answer is option c.

Note:

Keep in mind that 50=1 5^0=1

Answer

54 5^4

Exercise #11

10= 1^0=

Video Solution

Answer

1 1

Exercise #12

40=? 4^0=\text{?}

Video Solution

Answer

1 1

Exercise #13

(18)0=? (\frac{1}{8})^0=\text{?}

Video Solution

Answer

1

Exercise #14

(15)0= (\frac{1}{5})^0=

Video Solution

Answer

1 1

Exercise #15

00= 0^0=

Video Solution

Answer

Not defined

Topics learned in later sections

  1. Negative Exponents