Answer
The solution set is $\{1\}$
Work Step by Step
Use the hint, apply $\displaystyle \log_{a}M=\frac{\log_{b}M}{\log_{b}a}$
$\displaystyle \log_{4}x=\frac{\log_{2}x}{\log_{2}4}=\frac{\log_{2}x}{\log_{2}2^{2}}=\frac{\log_{2}x}{2}$
The equation becomes
$\displaystyle \log_{2}(x+1)-\frac{1}{2}\log_{2}x=\log_{2}2 \quad $... multiply with 2
Before continuing, the restrictions on x are
$\left\{\begin{array}{l}
x+1 \gt 0\\
x \gt 0
\end{array}\right.\Rightarrow x \gt 0\qquad(*)$
otherwise the equation is undefined (because of the logarithmic functions)
$2\log_{2}(x+1)-\log_{2}x=2\log_{2}2$ $\quad $...Apply$: \quad \log_{a}M^{r}=r\log_{a}M\quad $
$\log_{2}(x+1)^{2}-\log_{2}x=\log_{2}2^{2}$ $\quad $
...Apply$: \displaystyle \quad \log_{a}\left(\frac{M}{N}\right)=\log_{a}M-\log_{a}N$
$\displaystyle \log_{2}\left(\frac{(x+1)^{2}}{x}\right)=\log_{2}4 \quad $...Apply$: \quad $If $\log_{a}M=\log_{a}N,$ then $M=N$
$\displaystyle \frac{(x+1)^{2}}{x}=4 \quad $...multiply with x
$x^{2}+2x+1=4x$
$x^{2}-2x+1=0$
$(x-1)^{2}=0$
$ x=1\qquad$
so the solution set is $\{1\}$
