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$