( Logarithmic Equation ) Solve for x.

Question

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

Attempt at solution : $$ (x+1)^{log(x+1)} = 100(x+1)$$ $$= x^{log(x+1)} + 1 = 100x +1$$ $$=(x+1)+1=100x+1$$ $$=−98=99x$$ $$x=−98/99$$ But the answer given in the Practice sheet is :

$$= x=−9/10 or 99

Answer 1

Assuming $x>-1$, $$ (x+1)^{\log(x+1)}=100(x+1) $$ implies: $$ \log^2(x+1) = \log(100)+\log(x+1) $$ hence by setting $\log(x+1)=y$ we have: $$ y^2-y-2\log(10) = 0 $$ or: $$ y = \frac{1\pm\sqrt{1+8\log 10}}{2}$$ from which: $$ x=\exp\left(\frac{1\pm\sqrt{1+8\log 10}}{2}\right)-1. $$ If $\log$ is intended to be in base-$10$, the same process leads to $x=99$ or $x=-\frac{9}{10}$.

Answer 2

Here is my solution.I have left solving of quadratic equation to you