Multiplying and Dividing Exponents

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Math › Multiplying and Dividing Exponents

Questions 1 - 6

1

Simplify the following expression.

$\frac{2^{9}}{2^{11}}$

$\frac{1}{2^{2}}$

$2^{2}$

$\frac{18}{22}$

$\frac{9}{11}$

$2^{\frac{9}{11}}$

Explanation

Recall that when we are dividing exponents with the same base, we keep the base the same and subtract the exponents.

Thus, we have $\frac{2^{9}}{2^{11}} = 2^{9 - 11} = 2^{-2}$ .

We also recall that for negative exponents,

$b^{-x} = \frac{1}{b^{x}}$ .

Thus, $2^{-2} = \frac{1}{2^{2}}$ .

2

Simplify the expression:

$x^3 + \frac{2x^2}{x^{-1}}$

$x^3+\frac{2}{x^3}$

Explanation

First simplify the second term, and then combine the two:

$x^3 + \frac{2x^2}{x^{-1}}$

$= x^3 + 2x^{2-(-1)} = x^3 + 2x^3 = 3x^3$

3

Simplify the following exponent expression:

$(\frac{a^{\frac{-3}{2}}}{3^6b^{\frac{-2}{3}}})^{\frac{-1}{2}}$

$\frac{27a^{\frac{3}{4}}}{b^{\frac{1}{3}}}$

$\frac{18a^{\frac{3}{4}}}{b^{\frac{1}{3}}}$

$\frac{9a^{\frac{3}{4}}}{b^{\frac{1}{3}}}$

$\frac{3a^{\frac{3}{4}}}{b^{\frac{1}{3}}}$

$\frac{a^{\frac{3}{4}}}{b^{\frac{1}{3}}}$

Explanation

Begin by rearranging the terms in the numerator and denominator so that the exponents are positive:

$(\frac{a^{\frac{-3}{2}}}{3^6b^{\frac{-2}{3}}})^{\frac{-1}{2}}$

$(\frac{3^6b^{\frac{-2}{3}}}{a^{\frac{-3}{2}}})^{\frac{1}{2}}$

$(\frac{3^6a^{\frac{3}{2}}}{b^{\frac{2}{3}}})^{\frac{1}{2}}$

Multiply the exponents:

$(\frac{3^3a^{\frac{3}{4}}}{b^{\frac{1}{3}}})$

Simplify:

$\frac{27a^{\frac{3}{4}}}{b^{\frac{1}{3}}}$

4

Simplify the following expression.

$2^{3} \cdot 2^{6}$

$2^{9}$

$2^{18}$

$6^{6}$

$4^{18}$

$4^{9}$

Explanation

We are given: $2^{3} \cdot 2^{6}$ .

Recall that when we are multiplying exponents with the same base, we keep the base the same and add the exponents.

Thus, we have $2^{3 + 6} = 2^{9}$ .

5

Simplify the following expression.

$\frac{4x^3y^8z^2}{8x^6y^2z^4}$

$\frac{y^6}{2x^3z^2}$

$\frac{4y^6}{8x^3z^2}$

$\frac{4y^4}{8x^2z^2}$

$\frac{y^4}{2x^2z^2}$

Explanation

When dividing with exponents, the exponent in the denominator is subtracted from the exponent in the numerator. For example: $\frac{x^3}{x^6}= x^{3-6}=x^{-3}=\frac{1}{x^3}$ .

In our problem, each term can be treated in this manner. Remember that a negative exponent can be moved to the denominator.

$\frac{4x^3y^8z^2}{8x^6y^2z^4}=\frac{4}{8}x^{-3}y^6z^{-2}=\frac{4y^6}{8x^3z^2}$

Now, simplifly the numerals.

$\frac{4y^6}{8x^3z^2}=\frac{y^6}{2x^3z^2}$

6

Solve for :

Cannot be determined from the given information.

Explanation

Rewrite each side of the equation to only use a base 2:

$2^{2y} = 2^{5}*2^{2y-x}$

$2^{2y} = 2^{5+2y-x}$

The only way this equation can be true is if the exponents are equal.

So:

The on each side cancel, and moving the to the left side, we get:

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