Question

I need to use multi-step inverses with logs:

Answer 1

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

Step-by-step explanation:

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

Let f(x) be y

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

exchange x and y

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

simplify

`8x= {(7^y)^{(1)/(3) }+1}`

simplify

`8x - 1 = 7^{(y)/(3) }`

simplify

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

simplify

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

apply log rules

`y = (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)`