We have this equation:

First, combine both logarithms using the multiplication property and simplify the expression.
![\log[x(x + 99)] = 2](https://tex.z-dn.net/?f=%5Clog%5Bx%28x%20%2B%2099%29%5D%20%3D%202)
![\log[ {x}^{2} + 99x ] = 2](https://tex.z-dn.net/?f=%5Clog%5B%20%7Bx%7D%5E%7B2%7D%20%2B%2099x%20%5D%20%3D%202)
Now, use the definition of logarithm to transform the equation.


Finally, use the quadratic formula to solve the equation.

With this, we can say that the solution set is:
We cannot choose x = -100 as a solution because we cannot have a negative logarithm. The only solution is x = 1.
Answer:
11/6 as an improper fraction
Step-by-step explanation:
1 5/6 is 11/6 as an importer fraction and 1.86 as a decimal
The Y-Intercept would be on +4.
There’s no y or z within the problem though :/ only x