Question

I need to use multi-step inverses with logs:

Answer 1

Answer:

$f^{-1}(x)= \:\log _7\left(\left(8x-1\right)^3\right)$

Explanation:

$f(x) = \frac{(7^x)^{\frac{1}{3} }+1}{8}$

Let f(x) be y

$y = \frac{(7^x)^{\frac{1}{3} }+1}{8}$

exchange x and y

$x = \frac{(7^y)^{\frac{1}{3} }+1}{8}$

simplify

$8x= {(7^y)^{\frac{1}{3} }+1}$

simplify

$8x - 1 = 7^{\frac{y}{3} }$

simplify

$8x - 1 = \sqrt[3]{7^y}$

simplify

$(8x-1)^3 = 7^y$

apply log rules

$y = \frac{log((8x-1)^3 )}{log(7)}$

$y = \:\log _7\left(\left(8x-1\right)^3\right)$

therefore

$f^{-1}(x)= \:\log _7\left(\left(8x-1\right)^3\right)$