( Logarithmic Equation ) Solve for x. $(x+1)^{log(x+1)} = 100(x+1)$

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 :

$$= x=−9/10 or 99
Assuming $x&gt;-1$,
$$ (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 $x=-\frac{9}{10}$.
Here is my solution.I have left solving of quadratic equation to you