If log _{2}left(9^{x-1}+7right)-log _{2}left( | vedclass

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(D) Given: $\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2$ Using the property $\log_{a} m - \log_{a} n = \log_{a} (\frac{m}{n})$:

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$1, 2$

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(D) Given: $\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2$ Using the property $\log_{a} m - \log_{a} n = \log_{a} (\frac{m}{n})$: $\log _{2}\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=2$ Converting to exponential form: $\frac{9^{x-1}+7}{3^{x-1}+1}=2^{2}=4$ Let $y = 3^{x-1}$. Then $9^{x-1} = (3^{2})^{x-1} = (3^{x-1})^{2} = y^{2}$. Substituting $y$ into the equation: $\frac{y^{2}+7}{y+1}=4$ $y^{2}+7=4(y+1)$ $y^{2}-4y+3=0$ $(y-3)(y-1)=0$ So,$y=3$ or $y=1$. Case $1$: $3^{x-1}=3^{1}$ $\Rightarrow x-1=1$ $\Rightarrow x=2$. Case $2$: $3^{x-1}=3^{0}$ $\Rightarrow x-1=0$ $\Rightarrow x=1$. Thus,the values of $x$ are $1, 2$.

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

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