How to solve $ \log_8[\log_4(2x+1)] = \log_{27} 3$ for $x$?

Question

I'm stuck on this question.

$$\log_8[\log_4(2x+1)] = \log_{27} 3$$

What I did was to manipulate the inner $\log_4$ to get $\log_8[\frac{\log_2(2x+1)}{\log_2 4}] = \frac{1}{3}$

Then manipulating the outer $\log_8$ to become $\frac{\log_2 8}{\log_2 2}$, I will get $2x+1 = 2^{\frac{4}{9}}$, which is not what the answer key wanted. For reference, answer key gives $x = 7\frac{1}{2}$.

Greatly appreciate any hint. Thanks!

Answer 1

\begin{align} \log_8[\log_4(2x+1)] &= \log_{27} 3\\ \log_8[\log_4(2x+1)] &= \frac{1}{3}\\ \log_4(2x+1) &= 8^\frac{1}{3}=2\\ 2x+1&=4^2\\ x&=\frac{15}{2} \end{align}