# How do you solve for x in log(5-x) - 1/3log(35-x^3)=0?

Dec 6, 2015

Rearrange and derive a quadratic equation, one of whose roots is a valid solution of the original problem:

$x = \frac{75 - 3 \sqrt{105}}{26} \approx 1.702$

#### Explanation:

Add $\frac{1}{3} \log \left(35 - {x}^{3}\right)$ to both sides to get:

$\log \left(5 - x\right) = \frac{1}{3} \log \left(35 - {x}^{3}\right)$

Multiply both sides by $3$ to get:

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

Since $\log$ is one-one (as a Real valued function), we must have:

$35 - {x}^{3} = {\left(5 - x\right)}^{3} = {5}^{3} - 3 \left({5}^{2}\right) x + 3 \left(5\right) {x}^{2} - {x}^{3}$

$= 125 - 75 x + 13 {x}^{2} - {x}^{3}$

Add ${x}^{3}$ to both sides to get:

$13 {x}^{2} - 75 x + 125 = 35$

Subtract $35$ from both sides to get:

$13 {x}^{2} - 75 x + 90 = 0$

Use the quadratic formula to find:

$x = \frac{75 \pm \sqrt{{75}^{2} - 4 \cdot 13 \cdot 90}}{2 \cdot 13}$

$= \frac{75 \pm \sqrt{945}}{26}$

$= \frac{75 \pm 3 \sqrt{105}}{26}$

We need to check these solutions for validity:

If $x = \frac{75 + 3 \sqrt{105}}{26} \approx 4.067$ then ${x}^{3} > 64$, so $35 - {x}^{3} < 0$ and $\log \left(x\right)$ is not well defined.

If $x = \frac{75 - 3 \sqrt{105}}{26} \approx 1.702$ then $5 - x > 0$ and $35 - {x}^{3} \approx 35 - 4.93 > 0$, so both logs are well defined Real valued.