Question

How do you solve `log(x-15)=2-logx`?

Answer 1

`x = 20`

Explanation:

Put everything that's a log on the same side
`log(x-15)+log(x) = 2`

Remember that `log(m) + log(n) = log(mn)`
`log(x(x-15)) = 2`

If `log_a(b) = c`, then `b = a^c`
`x(x-15) = 10^(2)`

Expand and solve the quadratic equation

`x^2 - 15x = 100 rarr x^2 -15x -100 = 0`
`x = (15 +-sqrt(225 - 4*1(-100)))/2 = (15 +-sqrt(225 +400))/2`
`x = (15 +-sqrt(625))/2 = (15 +-25)/2`

`x_1 = (15+25)/2 = 40/2 = 20`
`x_2 = (15-25)/2 = -10/2 = -5`

Remember that since we were dealing with logarithms, we can't have null or negative arguments, so
`x-15 > 0 rarr x > 15`
`x > 0`

We conclude that any answers must follow `x > 15`, which only of the two answers do, thus, the answer is `x = 20`