Question

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

Answer 1

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`

Explanation:

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`

Use the quadratic formula to find:

`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.