If x+log _{10}left(1+2^{x}right)=x log _{10} | vedclass

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(C) Given equation: $x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6$Rearranging the terms:$\log _{10}\left(1+2^{x}\right) = x \log _{10}

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(C) Given equation: $x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6$Rearranging the terms:$\log _{10}\left(1+2^{x}\right) = x \log _{10} 5 - x + \log _{10} 6$$\log _{10}\left(1+2^{x}\right) = x \log _{10} 5 - x \log _{10} 10 + \log _{10} 6$$\log _{10}\left(1+2^{x}\right) = \log _{10} 5^{x} - \log _{10} 10^{x} + \log _{10} 6$$\log _{10}\left(1+2^{x}\right) = \log _{10}\left(\frac{5^{x} \cdot 6}{10^{x}}\right)$Taking antilog on both sides:$1+2^{x} = \frac{5^{x} \cdot 6}{2^{x} \cdot 5^{x}}$$1+2^{x} = \frac{6}{2^{x}}$Let $2^{x} = t$. Then $1+t = \frac{6}{t}$$t + t^{2} = 6$$t^{2} + t - 6 = 0$$(t+3)(t-2) = 0$Since $t = 2^{x} > 0$, we have $t = 2$.$2^{x} = 2^{1} \Rightarrow x = 1$.

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

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