You need to use the following rule of exponentiation, such that:

`a^(x+y) = a^x*a^y`

Reasoning by analogy, yields:

`x^(log x + 1) = x^(log x)*x^1`

Replacing `x*x^(log x)` for `x^(log x + 1)` yields:

`x*x^(log x) = 100`

Taking common logarithms both sides, yields:

`log (x*x^(log x)) = log 100`

Converting the logarithm of product into a sum of logarithms, yields:

`log x + log (x^(log x)) = log 10^2`

Using the power property of logarithms yields:

`log x + log x*log x = 2log 10`

Since `log 10 = ` 1 yields:

`log x + log^2 x = 2`

You should come up with the following substitution, such that:

`log x = y`

Replacing the variable yields:

`y + y^2 = 2 => y^2 + y - 2 = 0`

Using quadratic formula, yields:

`y_(1,2) = (-1+-sqrt(1+8))/2`

`y_(1,2) = (-1+-3)/2 => y_1 = 1; y_2 = -2`

You need to solve for x the following equations, such that:

`{(log x = 1),(log x = -2):} => {(x = 10),(x = 10^(-2)):}`

**Hence, evaluating the solutions to the given equation, yields **`x = 10, x = 1/100.`