If ${log}_{2} (3^{2 x + 2} + 7) = 2 + {log}_{2} (3^{x - 1} + 1)$ , then number of real values of $x$ is/are

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Solution

The correct option is A $0$
Given,
$l o g_{2} (3^{2 x + 2} + 7) = 2 + l o g_{2} (3^{x - 1} + 1)$

$\Rightarrow l o g_{2} (3^{2 x + 2} + 7) = l o g_{2} 4 + l o g_{2} (3^{x - 1} + 1)$

$= l o g_{2} (4) (3^{x - 1} + 1)$ $[∵ l o g a b = l o g a + l o g b]$

As both bases are same,

$\Rightarrow 3^{2 x + 2} + 7 = 4 (3^{x - 1} + 1)$

${9.3}^{2 x} - \frac{4}{3} {.3}^{x} + 6 = 0$

Let $3^{x} = u, \Rightarrow 9 u^{2} - \frac{4}{3} u + 6 = 0$

$27 u^{2} - 4 u + 18 = 0$

$\Rightarrow u = \frac{4 \pm \sqrt{16 - 4 (18) (27)}}{2 (27)}$

As $D = 16 - 4 (18) (27) < 0 \Rightarrow$ No solution exists

$∴$ Option A is correct.

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