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If $x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6$ then $x$ is equal to
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$\begin{aligned} & x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6 \\ & \Rightarrow x-x \log _{10} 5=\log _{10} 6-\log _{10}\left(1+2^{x}\right) \\ & \Rightarrow x\left(1-\log _{10} 5\right)=\log _{10} 6-\log _{10}\left(1+2^{x}\right) \\ & \Rightarrow x\left(\log _{10} 10-\log _{10} 5\right)=\log _{10}\left(\frac{6}{1+2^{x}}\right) \\ & \Rightarrow x\left(\log _{10}\left(\frac{10}{5}\right)\right)=\log _{10}\left(\frac{6}{1+2^{x}}\right) \\ \Rightarrow & x \log _{10} 2=\log _{10}\left(\frac{6}{1+2^{x}}\right) \end{aligned}$
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