Question

which of the following is equivalent to (3^x+1)(3^x-1)^2/9^x for all x? a. 3^2x-1 b. 3^x c. 3^x-1 d. 3^x-2 e. 3^3-x

Answer 1

Answer:

$3^{x-1}$

Explanation:

Given

$\frac{(3^{x+1})(3^{x-1})^2}{9^x}$

Required

Find its equivalent

$\frac{(3^{x+1})(3^{x-1})^2}{9^x}$

Expand the numerator

$\frac{(3^{x+1})(3^{x-1})(3^{x-1})}{9^x}$

Apply law of indices: $a^m * a^n = a^{m+n}$

This gives

$\frac{3^{x+1+x-1+x-1}}{9^x}$

Collect like terms

$\frac{3^{x+x+x+1-1-1}}{9^x}$

$\frac{3^{3x-1}}{9^x}$

Express $9^x$ as a factor of 3

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

Apply law of indices: $\frac{a^m}{a^n} = a^{m-n}$

This gives

$3^{3x-1-2x}$

$3^{3x-2x-1}$

$3^{x-1}$

Hence,

$\frac{(3^{x+1})(3^{x-1})^2}{9^x}$ is equivalent to $3^{x-1}$