Answer: \(x=15\) or \(x=-\frac12\).

Apply the laws of logarithms carefully, paying attention to the base of each logarithm and any restrictions on the argument.

Let

\(u=\log_2(x+1).\)

Then

\(\log_{(x+1)}2=\frac{1}{\log_2(x+1)}=\frac1u.\)

The equation becomes

\(u-\frac{4}{u}=3.\)

Multiply by \(u\):

\(u^2-4=3u.\)

So

\(u^2-3u-4=0.\)

Factorise:

\((u-4)(u+1)=0.\)

Hence \(u=4\) or \(u=-1\).

If \(\log_2(x+1)=4\), then \(x+1=16\), so \(x=15\).

If \(\log_2(x+1)=-1\), then \(x+1=\frac12\), so \(x=-\frac12\).

This gives the required answer: \(x=15\) or \(x=-\frac12\).