The number of solutions for the equation, ${(\log_{10} x)}^{2} + \log_{10} x^{2} = {(\log_{10} 2)}^{2} - 1$ is

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answer is B.

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Detailed Solution

{(\log_{10} x)}^{2} + \log_{10} x^{2} = {(\log_{10} 2)}^{2} - 1

$\log_{10} x = a$

$a^{2} + 2 a = {(\log_{10} 2)}^{2} - 1 \Rightarrow a^{2} + 2 a - {(\log_{10} 2)}^{2} + 1 = 0$

$= \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$= \frac{- 2 \pm \sqrt{4 - 4 ({(- \log_{10} 2)}^{2} + 1)}}{2}$

$= \frac{- 2 \pm \sqrt{4 + 4 {(\log_{10} 2)}^{2} - 4}}{2}$

$= \frac{- 2 \pm \sqrt{4 {(\log_{10} 2)}^{2}}}{2}$

$= \frac{- 2 \pm 2 \log_{10} 2}{2}$

$= \frac{2 (- 1 \pm \log_{10} 2)}{2}$

$= - 1 \pm \log_{10} 2$

$∴$ The number of solutions is 2.

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