If \(\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2\), then \(x\) values are

\(\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2\)
\(\Rightarrow \quad \log _{2}\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=2 \log _{2} 2\)
\(\Rightarrow \quad \log _{2}\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=\log _{2} 2^{2}\)
\(\Rightarrow \quad\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=4\)
\(\Rightarrow \quad \frac{\left(3^{2}\right)^{x-1}+7}{\left(3^{x-1}+1\right)}=4\)
\(\Rightarrow \quad \frac{\left(3^{x-1}\right)^{2}+7}{3^{x-1}+1}=4\)
Let \(\quad 3^{\mathrm{x}-1}=\mathrm{y}\)
\(\therefore \quad \frac{y^{2}+7}{y+1}=4\)
\(\Rightarrow \quad \mathrm{y}^{2}+7=4 \mathrm{y}+4\)
\(\Rightarrow \quad y^{2}-4 y+3=0\)
\(\Rightarrow \quad y^{2}-3 y-y+3=0\)
\(\Rightarrow y(y-3)-1(y-3)=0\)
\(\Rightarrow \quad(\mathrm{y}-3)(\mathrm{y}-1)=0\)
\(\Rightarrow \quad y=3,1\)
If \(y=3\), then
\(3^{x-1}=3\)
\(\Rightarrow \quad x-1=1 \Rightarrow x=2\)
If \(\mathrm{y}=1\), then
\(3^{x-1}=3^{0}\)
\(\Rightarrow \quad x-1=0 \Rightarrow x=1\)
\(\therefore \quad x=1,2\)