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

Question

2315 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-06T11:24:28

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

$x = (75-3sqrt(105))/26 ~~ 1.702$

Add $1/3 log(35-x^3)$ to both sides to get:

$log(5-x) = 1/3 log(35-x^3)$

Multiply both sides by $3$ to get:

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

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

$35-x^3 = (5-x)^3 = 5^3-3(5^2)x+3(5)x^2-x^3$

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

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

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

Subtract $35$ from both sides to get:

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

$x = (75+-sqrt(75^2-4*13*90))/(2*13)$

$=(75+-sqrt(945))/26$

$=(75+-3sqrt(105))/26$

We need to check these solutions for validity:

If $x = (75+3sqrt(105))/26 ~~ 4.067$ then $x^3 > 64$ , so $35-x^3 < 0$ and $log(x)$ is not well defined.

If $x = (75-3sqrt(105))/26 ~~ 1.702$ then $5-x > 0$ and $35-x^3 ~~ 35-4.93 > 0$ , so both logs are well defined Real valued.

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