If $f(x) =\frac{3 + 7^{2x}}4$, which of the following is the inverse? If $f(x) = \frac{3 + 7^{2x}}  4$, what is the inverse function $f^{-1}(x)$?
Initially, I thought it would simplify down to $\frac 12 \log_7(4x-3)$
But apparently that is not correct.
 A: Your answer seems correct to me. Let me write this fact with different notations:
$$f^{-1}(x)=\log_{49}(4x-3), x>\frac 34\\
f^{-1}(x)=\frac 12 \log_7(4x-3), x>\frac 34\\
f^{-1}(x)=\frac {\ln(4x-3)}{\ln 49}, x>\frac 34\\
f^{-1}(x)=\frac {\ln(4x-3)}{2\ln 7}, x>\frac 34$$
Hopefully, one of these will be among the answers.  That all these expressions are equal can be easily proved by simple logarithm rules.
A: Note that $f: \mathbb R \to (3/4, \infty)$ is continuous and strictly increasing therefore it is invertible, that is
$$\begin{aligned}y =\frac{3 + 7^{2x}}4 &\iff 4y=3 + 7^{2x} \\
&\iff 4y-3 =7^{2x} \\
&\iff \log_7 (4y-3)=2x\\
&\iff x=\frac12 \log_7 (4y-3)\end{aligned}$$
and the inverse function is
$$g(x)=f^{-1}(x)=\frac12 \log_7 (4x-3)$$
which corresponds to your guess. Maybe possible answers are given in an equivalent form. To check you can plug $f(x)$ in the inverse to obtain $x$ as result, that is
$$\frac12 \log_7 \left(4\frac{3 + 7^{2x}}4-3\right) = \frac12\log_7 (7^{2x})= \frac12 2x=x$$
