Question

If ${\log }_{10}\left(\frac{1}{2^x+x-1}\right)=x({\log }_{10}5-1)$, then the value of $x$will be

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

$1$

${\log }_{10}\left(\frac{1}{2^x+x-1}\right)=x({\log }_{10}5-1)\Rightarrow -{\log }_{10}(2^x+x-1)[\because \log a^m=m\log a]=x({\log }_{10}5-{\log }_{10}10)[\because {\log }_{a}a=1]\Rightarrow {\log }_{10}(2^x+x-1)=-x{\log }_{10}\frac{5}{10}[\because \log a-\log b=\log \frac{a}{b}]\Rightarrow {\log }_{10}(2^x+x-1)=x{\log }_{10}2\Rightarrow {\log }_{10}(2^x+x-1)={\log }_{10}{2}^x\Rightarrow 2^x+x-1=2^x\Rightarrow x=1$
Ans: D