Question

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

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

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

$=x\left[{\log }_{10}5-{\log }_{10}10\right]$

$=x\left[{\log }_{10}\frac{5}{10}\right]$

$={\log }_{10}{\left(\frac{1}{2}\right)}^x$

$\therefore {\log }_{10}\left[\frac{1}{2^x+x-1}\right]={\log }_{10}{\left(\frac{1}{2}\right)}^x$

$2^x+x-1=2^x$
$x-1=0$
$x=1$