Question

If ${\log }_{10}\left[\frac{1}{2^x+x-1}\right]=x[{\log }_{10}5-1]$, then $x$ equals to

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

${\log }_{10}[\frac{1}{2^x+x-1}]=x[{\log }_{10}5-1]\text{Now,}{2}^x+x-1>0\Rightarrow x>0\text{R.H.S}=x({\log }_{10}5-1)=x({\log }_{10}5-{\log }_{10}10)=x{\log }_{10}\left(2^{-1}\right)={\log }_{10}(\frac{1}{2^x})\because \text{L.H.S = R.H.S}\therefore \frac{1}{2^x+x-1}=\frac{1}{2^x}\Rightarrow x-1=0\Rightarrow x=1$