If $x + {log}_{10} (1 + 2^{x}) = x {log}_{10} 5 + {log}_{10} 6$ , then the value of $x$ is

\frac{1}{2}

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\frac{1}{3}

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1

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2

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Solution

The correct option is D $1$
${log}_{10} (1 + 2^{x}) = {log}_{10} 5^{x} + {log}_{10} 6^{x} - {log}_{10} 10^{x}$

$1 + 2^{x} = \frac{6 \times 5^{x}}{10^{x}} = \frac{6}{2^{x}}$

let $2^{x} = t$

$t (1 + t) = 6$
$\Rightarrow t^{2} + t - 6 = 0$

$(t + 3) (t - 2) = 0$

$2^{x} = 2$

$\Rightarrow x = 1$

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