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

instagram story viewer

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:

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

Check out some free courses

Free Online Inclusive Education Course
Free Online Toy Library and Learning Course
Free Online Math Games Course in Early Childhood Education
Free Online Pedagogical Cultural Workshops Course

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

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

You may also be interested:

List of Radiation Exercises
Logarithm Exercise List
List of Numerical Expression Exercises

The password has been sent to your email.

Exercises on properties of potencies

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

Cheers with the letter M

Writing: How to Teach Students to Write Well

The semantic relationships between words