Alg9.4.50

If $\log_{10}(\frac{1}{2^x+x-1})=x(\log_{10}(5)-1)\,$ then find the value of x.

Simplifying the right hand side expression,

$x(\log_{10} 5-\log_{10} (10))\,$

$x(\log_{10}(\frac{5}{10})\,$

$\log_{10}(\frac{1}{2})^{x}\,$

$\log_{10}(\frac{1}{2^x})\,$

Equating with the lefthand side expression and removing the logs on both sides,

we get

$\frac{1}{2^x+x-1}=\frac{1}{2^x}\,$

Simplifying this equation,we get

$x-1=0\,$

Hence the value of $x\,$ is 1.