If ${log}_{2} (3^{2 x - 2} + 7) = 2 + {log}_{2} (3^{x - 1} + 1)$ , then value of $x$ is equal to

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Solution

The correct options are
B $1$
C $2$
${log}_{2} (3^{2 x - 2} + 7) = 2 + {log}_{2} (3^{x - 1} + 1)$
${log}_{2} (3^{2 x - 2} + 7) = 2 {log}_{2} 2 + {log}_{2} (3^{x - 1} + 1) [∵ {log}_{a} a = 1]$
$\Rightarrow {log}_{2} (3^{2 x - 2} + 7) = {log}_{2} 4 (3^{x - 1} + 1) [∵ m log a = log a^{m} & log a + log b = log (a b)]$
$\Rightarrow 3^{2 x - 2} + 7 = 4 (3^{x - 1} + 1)$
$\Rightarrow {(3^{x - 1})}^{2} - 4 (3^{x - 1}) + 3 = 0$
$\Rightarrow (3^{x - 1} - 1) (3^{x - 1} - 3) = 0$
$\Rightarrow 3^{x - 1} = 1, 3^{x - 1} = 3$
$\Rightarrow x - 1 = 0, x - 1 = 1$
$\Rightarrow x = 1, 2$
Ans: B,C

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