If log _{2}left(3^{2 x-2}+7right)=2+log _{2}l | vedclass

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(D) Given equation: $\log _{2}\left(3^{2 x-2}+7\right)=2+\log _{2}\left(3^{x-1}+1\right)$Using the property $2 = \log _{2} 4$,we get:$\log _{2}\left(3

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(D) Given equation: $\log _{2}\left(3^{2 x-2}+7\right)=2+\log _{2}\left(3^{x-1}+1\right)$Using the property $2 = \log _{2} 4$,we get:$\log _{2}\left(3^{2 x-2}+7\right)=\log _{2} 4+\log _{2}\left(3^{x-1}+1\right)$$\log _{2}\left(3^{2 x-2}+7\right)=\log _{2}\left[4\left(3^{x-1}+1\right)\right]$Removing the logarithm from both sides:$3^{2 x-2}+7=4\left(3^{x-1}+1\right)$Let $t = 3^{x-1}$. Then $3^{2x-2} = (3^{x-1})^2 = t^2$.The equation becomes: $t^2 + 7 = 4(t + 1)$$t^2 + 7 = 4t + 4$$t^2 - 4t + 3 = 0$$(t - 1)(t - 3) = 0$So,$t = 1$ or $t = 3$.Case $1$: $t = 1 \Rightarrow 3^{x-1} = 1 = 3^0 \Rightarrow x - 1 = 0 \Rightarrow x = 1$.Case $2$: $t = 3 \Rightarrow 3^{x-1} = 3^1 \Rightarrow x - 1 = 1 \Rightarrow x = 2$.Thus,$x = 1$ or $x = 2$.

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

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