The problem asks to solve the equation $\frac{\log(35-x^3)}{\log(5-x)} = 3$.

1. Problem Description

The problem asks to solve the equation $\frac{\log(35-x^3)}{\log(5-x)} = 3$.

2. Solution Steps

We are given the equation:

$\frac{\log(35-x^3)}{\log(5-x)} = 3$

Using the change of base formula for logarithms, we can rewrite this as:

$\log_{5-x}(35-x^3) = 3$

This means that:

$(5-x)^3 = 35-x^3$

Expanding the left side, we have:

$125 - 75x + 15x^2 - x^3 = 35 - x^3$

Simplifying the equation:

$125 - 75x + 15x^2 - x^3 + x^3 - 35 = 0$

$15x^2 - 75x + 90 = 0$

Divide by 15:

$x^2 - 5x + 6 = 0$

Factoring the quadratic:

$(x-2)(x-3) = 0$

Therefore, the possible solutions are $x=2$ and $x=3$.

Now, we need to check for extraneous solutions. The logarithm is only defined for positive arguments, and the base must be positive and not equal to

1. Thus:

$5-x > 0$, which implies $x < 5$.

$5-x \neq 1$, which implies $x \neq 4$.

$35 - x^3 > 0$, which implies $x^3 < 35$, or $x < \sqrt[3]{35} \approx 3.27$.

If $x=2$, then $5-x = 5-2 = 3 > 0$ and $5-x \neq 1$. Also $35-x^3 = 35 - 2^3 = 35-8 = 27 > 0$. Thus, $x=2$ is a valid solution.

If $x=3$, then $5-x = 5-3 = 2 > 0$ and $5-x \neq 1$. Also $35-x^3 = 35 - 3^3 = 35-27 = 8 > 0$. Thus, $x=3$ is a valid solution.

2. Final Answer

The solutions are $x=2$ and $x=3$.