Question

Solve for $x$

${(x+1)}^{\log (x+1)}=100(x+1)$

Answer 1

${(x+1)}^{\log (x+1)}=100(x+1)$

Applying $\log$ on both sides, we get

$\log \left({(x+1)}^{\log (x+1)}\right)=\log \left(100(x+1)\right)$

$\Rightarrow {\log }_{10}(x+1){\log }_{10}(x+1)={\log }_{10}100+{\log }_{10}(x+1)$

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

$\Rightarrow {\left({\log }_{10}(x+1)\right)}^2-{\log }_{10}(x+1)-2=0$

Let $t={\log }_{10}(x+1)$

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

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

$\Rightarrow t(t-2)+1(t-2)=0$

$\Rightarrow (t-2)(t-1)=0$

$\therefore t=1,2$

$\Rightarrow {\log }_{10}(x+1)=1,{\log }_{10}(x+1)=2$

$\Rightarrow x+1=10,x+1={10}^2=100$

$\Rightarrow x=10-1=9,x=100-1=99$

$\therefore x=9,x=99$