# Solve in equation for y=log(15-x)(x+5) y=f(10-x)?base log 15-x

*print*Print*list*Cite

### 1 Answer

You need to solve the equation f(x)=f(10-x), hence you need to plug 10 -x in the equation of function such that:

`log_(15 - x) (x + 5) = log_(15 - 10 + x) (10 - x + 5)`

`log_(15 - x) (x + 5) = log_(5 + x) (15 - x)`

You need to convert `log_(15 - x) (x + 5)` to the base (x + 5) such that:

`log_(15 - x) (x + 5) = 1/(log_(5 + x) (15 - x))`

Rewriting the equation yields:

`1/(log_(5 + x) (15 - x)) = log_(5 + x) (15 - x)`

You should come up with the substitution `log_(5 + x) (15 - x) = y` such that:

`1/y = y =gt 1 = y^2 =gt y_(1,2) = +-1`

Hence, `log_(5 + x) (15 - x) = +-1`

`log_(5 + x) (15 - x) = 1`

Going in reverse yields:

`15 - x = (x + 5)^1=gt 15 - x = x + 5`

`` Adding x both sides yields:

`15 = 2x + 5 =gt 2x = 15 - 5 =gt 2x = 10 =gt x = 5`

`log_(5 + x) (15 - x) = -1 =gt 15 - x = (x+5)^(-1)`

`` `15 - x = 1/(x + 5)`

You need to bring to a common denominator such that:

`(15 - x)(x + 5) - 1 = 0`

`` `15x + 75 - x^2 - 5x - 1 = 0`

`- x^2 + 10x + 74 = 0=gt x^2 - 10x - 74 = 0`

`x_(1,2) = (10 +- sqrt(100 + 296))/2 =gt x_1 ~~14`

`` `x_2~~ 5`

`` **Notice that solutions to equation need to be in the interval (-5 , 15), hence, the solutions to the given equation are:`x_1 = 5` ; `x_2~~ 14` .**