The problem is to solve the equation $(x+1)^{\log(x+1)} = 100(x+1)$. It is assumed the base of the logarithm is 10.

1. Problem Description

The problem is to solve the equation $(x+1)^{\log(x+1)} = 100(x+1)$. It is assumed the base of the logarithm is

1

0.

2. Solution Steps

We have the equation $(x+1)^{\log(x+1)} = 100(x+1)$.

Take the base 10 logarithm of both sides:

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

Using the logarithm property $\log(a^b) = b\log(a)$, we have

$\log(x+1)\cdot \log(x+1) = \log(100) + \log(x+1)$

Also $\log(100) = \log(10^2) = 2$. So the equation becomes

$(\log(x+1))^2 = 2 + \log(x+1)$

Let $y = \log(x+1)$. The equation is now

$y^2 = 2+y$

$y^2 - y - 2 = 0$

$(y-2)(y+1) = 0$

So, $y = 2$ or $y = -1$.

Case 1: $y = 2$.

$\log(x+1) = 2$

$x+1 = 10^2 = 100$

$x = 100 - 1 = 99$.

Case 2: $y = -1$.

$\log(x+1) = -1$

$x+1 = 10^{-1} = \frac{1}{10}$

$x = \frac{1}{10} - 1 = \frac{1}{10} - \frac{10}{10} = -\frac{9}{10}$.

We need to check the validity of solutions. For $x=99$, we have $x+1 = 100$, so $(x+1)^{\log(x+1)} = 100^{\log(100)} = 100^2 = 10000$. On the other hand, $100(x+1) = 100(100) = 10000$. So $x=99$ is a valid solution.

For $x=-\frac{9}{10}$, we have $x+1 = \frac{1}{10}$, so $(x+1)^{\log(x+1)} = (\frac{1}{10})^{\log(\frac{1}{10})} = (\frac{1}{10})^{-1} = 10$. On the other hand, $100(x+1) = 100(\frac{1}{10}) = 10$. So $x=-\frac{9}{10}$ is also a valid solution.

3. Final Answer

$x=99$ or $x=-\frac{9}{10}$