Exercises on properties of potencies

THE potentiation is a mathematical operation used to express the product of a number by itself. This operation has some important properties, which make it possible to simplify and solve many calculations.

The main potentiation properties they are:

→ Potentiation with an exponent equal to zero:

$\dpi{120} \mathbf{a^0 = 1, a\neq 0}$

→ Potentiation with an exponent equal to 1:

$\dpi{120} \mathbf{a^1 = a}$

→ Potentiation of negative numbers with $\dpi{120} \mathrm{a>0}$ and $\dpi{120} \mathrm{m}$ an even number:

$\dpi{120} \mathbf{(-a)^m = a^m}$

→ Potentiation of negative numbers with $\dpi{120} \mathrm{a>0}$ and $\dpi{120} \mathrm{m}$ an odd number:

$\dpi{120} \mathbf{(-a)^m = -(a^m) }$

→ Power of a power:

$\dpi{120} \mathbf{(a^m)^n = a^{m\cdot n}}$

→ Power with negative exponent:

$\mathbf{a^{-m} = \bigg(\frac{1}{a}\bigg)^m = \frac{1}{a^m}}$

→ Power multiplication:

$\dpi{120} \mathbf{a^m\cdot a^n = a^{m+n}}$

→ Power division:

$\dpi{120} \mathbf{a^m: a^n = a^{m-n}}$

To learn more, check out a list of exercises on potency properties. All issues resolved for you to clear your doubts.

Index

Exercises on properties of potencies
Resolution of question 1
Resolution of question 2
Resolution of question 3
Resolution of question 4
Resolution of question 5
Resolution of question 6
Resolution of question 7
Resolution of question 8

Exercises on properties of potencies

Question 1. Calculate the following powers: $\dpi{120} (-3)^2$ , $\dpi{120} (-1)^9$ , $\dpi{120} (-5)^3$ and $\dpi{120} (-2)^6$ .

Question 2. Calculate the following powers: $\dpi{120} 4^2$ , $\dpi{120} -4^2$ and $\dpi{120} (-4)^2$ .

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Question 3. Calculate the negative exponent powers: $\dpi{120} 5^{-1}$ , $\dpi{120} 8^{-2}$ , $\dpi{120} (-3)^{-3}$ and $\dpi{120} (-1)^{-8}$ .

Question 4. Calculate the following powers: $\dpi{120} (4^2)^3$ , $\dpi{120} (-2^3)^{-1}$ , $\dpi{120} (3^2)^{-2}$ and $\dpi{120} (5^{-1})^{-2}$ .

Question 5. Make the multiplications between powers:

$\dpi{120} 3^2\cdot 3^3$

$\dpi{120} 2^2\cdot 2^{-2}\cdot 2^{3}$

$\dpi{120} 3^{-1}\cdot 5^5\cdot 3^2\cdot 5^{-3}\cdot 5^1$

Question 6. Make the divisions between powers: $\dpi{120} \frac{3^6}{3^4}$ , $\dpi{120} \frac{2^5}{2^0}$ and $\dpi{120} \frac{5^{-9}}{5^{-7}}$ .

Question 7. Calculate the following powers: $\dpi{120} \left ( \frac{2}{3} \right )^2$ , $\dpi{120} \left ( -\frac{2}{5} \right )^3$ , $\dpi{120} \left ( \frac{5}{2} \right )^4$ .

Question 8. Calculate:

$\dpi{120} \frac{2^3\cdot 3^{-2}\cdot 2^0\cdot 2^{-5}\cdot 3^1}{3^3\cdot 2^5\cdot 3 ^{-2}}$

Resolution of question 1

As in $\dpi{120} (-3)^2$ the exponent is even, the power will be positive:

$\dpi{120} (-3)^2 = 3^2 = 9$

As in $\dpi{120} (-1)^9$ the exponent is odd, the power will be negative:

$\dpi{120} (-1)^9 = -(1^9) = -1$

As in $\dpi{120} (-5)^3$ the exponent is odd, the power will be negative:

$\dpi{120} (-5)^3 = -(5^3)= - 125$

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As in $\dpi{120} (-2)^6$ the exponent is even, the power will be positive:

$\dpi{120} (-2)^6= 2^6 = 64$

Resolution of question 2

In all three cases, the power will be the same, except for the sign, which can be positive or negative:

$\dpi{120} 4^2 = 16$

$\dpi{120} -4^2 =- (4^2) = -16$

$\dpi{120} (-4)^2 = 4^2 = 16$

Resolution of question 3

the power $\dpi{120} 5^{-1}$ is the inverse of power $\dpi{120} 5^{1}$ :

$\dpi{120} 5^{-1} = \frac{1}{5^1} = \frac{1}{5}$

the power $\dpi{120} 8^{-2}$ is the inverse of power $\dpi{120} 8^{2}$ :

$\dpi{120} 8^{-2} = \frac{1}{8^2} = \frac{1}{64}$

the power $\dpi{120} (-3)^{-3}$ is the inverse of power $\dpi{120} (-3)^{3}$ :

$\dpi{120} (-3)^{-3} = \frac{1}{(-3)^3} = \frac{1}{-(3^3)} = -\frac{1}{ 27}$

the power $\dpi{120} (-1)^{-8}$ is the inverse of power $\dpi{120} (-1)^{8}$ :

$\dpi{120} (-1)^{-8} = \frac{1}{(-1)^8} = \frac{1}{1^8} = 1$

Resolution of question 4

In each case, we can multiply the exponents and then calculate the power:

$\dpi{120} (4^2)^3 = 4^{2\cdot 3} = 4^6 = 4096$

$\dpi{120} (-2^3)^{-1} =(-2)^{3\cdot -1} = (-2)^{-3} = \frac{1}{(-2) ^3} = -\frac{1}{8}$

$\dpi{120} (3^2)^{-2} = 3^{2\cdot -2} = 3^{-4} = \frac{1}{3^4} = \frac{1}{ 81}$

$\dpi{120} (5^{-1})^{-2} = 5^{-1\cdot -2} = 5^2 = 25$

Resolution of question 5

In each case, we add the exponents of the powers of the same base:

$\dpi{120} 3^2\cdot 3^3 = 3^{2 + 3} = 3^5= 243$

$\dpi{120} 2^2\cdot 2^{-2}\cdot 2^{3} = 2^{2 -2 +3} = 2^3 = 8$

$\dpi{120} 3^{-1}\cdot 5^5\cdot 3^2\cdot 5^{-3}\cdot 5^1 = 3^{-1 +2}\cdot 5^{5- 3+1}= 3^1\cdot 5^3 = 3\cdot 125 = 375$

Resolution of question 6

In each case, we subtract the exponents of the powers of the same base:

$\dpi{120} \frac{3^6}{3^4}= 3^{6 -4} = 3^2 =9$

$\dpi{120} \frac{2^5}{2^0} = 2^{5-0} =2^5 = 32$

$\dpi{120} \frac{5^{-9}}{5^{-7}} = 5^{-9 -(-7)} = 5^{-9+7} = 5^{-2 }= \frac{1}{25}$

Resolution of question 7

In each case, we raise both terms to the exponent:

$\dpi{120} \left ( \frac{2}{3} \right )^2 = \frac{2^2}{3^3} = \frac{4}{27}$

$\dpi{120} \left ( -\frac{2}{5} \right )^3 = -\frac{2^3}{5^3} = -\frac{8}{125}$

$\dpi{120} \left ( \frac{5}{2} \right )^4 = \frac{5^4}{2^4} = \frac{625}{16}$

Resolution of question 8

$\dpi{120} \small \frac{2^3\cdot 3^{-2}\cdot 2^0\cdot 2^{-5}\cdot 3^1}{3^3\cdot 2^5\ cdot 3^{-2}} = \frac{2^{-2}\cdot 3^{-1}}{3^{1}\cdot 2^5} = 2^{-2-5}\cdot 3^{-1-1} = 2^{-7}\cdot 3^{-2} = \frac{1}{2^7\cdot 3^2} = \frac{1}{1152}$