Question

Find the value of $x$ which satisfies, $x+{\log }_{10}(1+2^x)=x{\log }_{10}5+{\log }_{10}6$.

Answer 1

Given $x+{\log }_{10}(1+2^x)=x{\log }_{10}5+{\log }_{10}6$

$\Rightarrow x{\log }_{10}10+{\log }_{10}(1+2^x)=x{\log }_{10}5+{\log }_{10}6\dots (lo{g}_{a}a=1)$

We know that ${\log }_{a}m+{\log }_{a}n={\log }_{a}mn$

${\log }_a{m}^n=n{\log }_{a}m$

$\Rightarrow {\log }_{10}{10}^x+{\log }_{10}(1+2^x)={\log }_{10}{5}^x+{\log }_{10}6$

$\Rightarrow {10}^x(1+2^x)=6\cdot 5^x$

$\Rightarrow 2^x(1+2^x)=6$

$\Rightarrow 2^x+2^{2x}-6=0$

Let $2^x=t$

$\Rightarrow t^2+t-6=0$

$\Rightarrow (t+3)(t-2)=0$

After solving the above equation, we get

$2^x=-3$ (not possible)

Or $2^x=2$

$\Rightarrow x=1$