Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. 1
• D. 2

Answer 1

Answer: (C)

We have,
$x+{\log }_{10}\left(1+2^x\right)=x{\log }_{10}5+{\log }_{10}6$
$\Rightarrow {\log }_{10}\left(1+2^x\right)=x{\log }_{10}5+{\log }_{10}6-x$
$={\log }_{10}{5}^x+{\log }_{10}6-x{\log }_{10}10$
$={\log }_{10}\left(5^x\cdot 6\right)-{\log }_{10}10^x$
$\Rightarrow {\log }_{10}\left(1+2^x\right)={\log }_{10}\left(\frac{5^x\cdot 6}{10^x}\right)$
$\Rightarrow 1+2^x=\frac{5^x\cdot 6}{10^x}=\frac{6}{2^x}$
$\Rightarrow 2^x\left(1+2^x\right)=6$
$\Rightarrow t(1+t)=6\left(\text{ let }{2}^x=t\right)$
$\Rightarrow (t+3)(t-2)=0$
$t+3=0$
or $t-2=0$
$\Rightarrow t=2[\because$ neglect $t=3]$
$\Rightarrow 2^x=2$
$\Rightarrow x=1$