Question

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

Answer 1

`S={20}`

Explanation:

Use the log property that `log(a)+log(b)=log(ab)`, so

`log(x-15)+log(x)=2`
`log(x(x-15))=2`

Since no base was specified we assume the base of the logarithm was `10`, take the base-10 exponential of both sides, i.e.:

`10^(log(x(x-15)))=10^2`

We know that `b^(log_b(a))=a`, and that `10^2=100` so

`(x(x-15))=100`

Expand the product and take that 100 to the LHS

`x^2-15x-100=0`

Solve the quadratic in your favorite way, by the quadratic formula for example

`x=(15+-sqrt(15^2-4*1*(-100)))/(2*1)=(15+-sqrt(225+400))/(2)`
`x=(15+-sqrt(625))/(2)=(15+-25)/(2)`
`x^'=(15+25)/(2)=(40)/(2)=20`
`x^'=(15-25)/(2)=(-10)/(2)=-5`

However, recall that this was originally a logarithmic formula; we can have anything equal to 0 or a negative number in one!

So we know that

`x-15>0`
`x>0`

`20-15=5>0` and `20>0` so it's a valid solution, but `-5<0` so it isn't one. So the set of solutions is `S={20}`