Solve for x :x+ _ 10 (1+2^x )=x _ 10 5+ _ 10 6

Question

Solve for $\mathrm{x}$ :$x+\log _{10}\left(1+2^x\right)=x \log _{10} 5+\log _{10} 6$

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Answer

$ x+\log _{10}\left(1+2^x\right)=x \log _{10} 5+\log _{10} 6$
$\therefore \quad x \log _{10} 10+\log _{10}\left(1+2^x\right)=x \log _{10} 5+\log _{10} 6$
$.\left[\log _a a=1\right]$
$\therefore \quad \log _{10} 10^x+\log _{10}\left(1+2^x\right)=\log _{10} 5^x+\log _{10} 6$
$\ldots\left[n \log m=\log \mathrm{m}^{\mathrm{n}}\right]$
$\therefore \quad \log _{10}\left[10^x \cdot\left(1+2^x\right)\right]=\log _{10}\left(6 \times 5^x\right)$
$\ldots[\log m+\log n=\log m n]$
$\therefore \quad 10^x\left(1+2^x\right)=6 \times 5^x$
$\therefore \quad 2^x \times 5^x\left(1+2^x\right)=6 \times 5^x$
$\therefore \quad 2^x\left(1+2^x\right)=6$
$\text { Let } 2^x=a$
$\therefore \quad a \cdot(1+a)=6$
$\therefore a+a^2=6$
$\therefore a^2+a-6=0$
$\therefore(a+3)(a-2)=0$
$\therefore a+3=0 \text { or } a-2=0$
$\therefore \mathrm{a}=-3 \text { or } \mathrm{a}=2$
$\text { Since } 2 x=-3 \text { is not possible, }$
$2 x=2=21$
$\therefore x=1$

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