UPSC NDA 2 2017 Mathematics Solved Paper

Concept:Use logarithm properties to combine terms and simplify the equation, then convert to an exponential form and solve for $x$.Explanation:Start with: $x + \log_{10}(1+2^{x}) = x \log_{10}5 + \log_{10}6$.Bring logarithmic terms to one side: $x - x \log_{10}5 = \log_{10}6 - \log_{10}(1+2^{x})$.Factor $x$: $x(1 - \log_{10}5) = \log_{10}6 - \log_{10}(1+2^{x})$.Since $1 = \log_{10}10$, rewrite: $x(\log_{10}10 - \log_{10}5) = \log_{10}\left(\frac{6}{1+2^{x}}\right)$.The left side simplifies: $\log_{10}10 - \log_{10}5 = \log_{10}\left(\frac{10}{5}\right) = \log_{10}2$. So we have $x \log_{10}2 = \log_{10}\left(\frac{6}{1+2^{x}}\right)$.Apply the power rule of logarithms: $\log_{10}(2^{x}) = \log_{10}\left(\frac{6}{1+2^{x}}\right)$.Since the bases are equal, equate the arguments: $2^{x} = \frac{6}{1+2^{x}}$.Multiply both sides by $(1+2^{x})$: $2^{x}(1+2^{x}) = 6$.Let $u = 2^{x}$. Then $u(1+u) = 6$, i.e., $u^{2} + u - 6 = 0$.Factor: $(u+3)(u-2) = 0$, so $u = -3$ or $u = 2$.Since $2^{x} > 0$ for any real $x$, $u = -3$ is invalid. Hence $2^{x} = 2$, giving $x = 1$.Answer:$x = 1$ (option C).