Answer:
f-1(x) = log2 [ (x + 4)/3 ] - 3.
Step-by-step explanation:
Let y = 3(2^(x+3) - 4
3(2^(x + 3) = y + 4
2^(x + 3) = (y + 4)/3
x + 3 = log2 [(y + 4)/3]
x = log2 [ (y + 4)/3 ] - 3
Inverse f-1(x) = log2 [ (x + 4)/3 ] - 3
as you already know, we start by doing a quick switcheroo on the variables, to get the inverse expression of any expression, so
[tex]\bf \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{f(x)}{y}=3\left( 2^{x+3} \right)-4\implies \stackrel{\textit{quick switcheroo}}{x=3\left( 2^{y+3} \right)-4}\implies x+4=3\left( 2^{y+3} \right)[/tex]
[tex]\bf \cfrac{x+4}{3}=2^{y+3}\implies \log\left( \cfrac{x+4}{3} \right)=\log\left( 2^{y+3} \right) \\\\\\ \log\left( \cfrac{x+4}{3} \right)=(y+3)\log\left( 2 \right)\implies \cfrac{\log\left( \frac{x+4}{3} \right)}{\log(2)}=y+3 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{\log\left( \frac{x+4}{3} \right)}{\log(2)}-3=\stackrel{f^{-1}(x)}{y}~\hfill[/tex]
