Question

How do you solve `log x + log (x-3) = 1`?

Answer 1

`x = 5`

Explanation:

Assuming `log` is referring to the base-`10` logarithm, we can apply the properties that `log(a)+log(b) = log(ab)` and `10^log(x)=x` to obtain

`log(x)+log(x-3)=1`

`=> log(x(x-3)) = 1`

`=> 10^log(x(x-3)) = 10^1`

`=> x(x-3) = 10`

`=> x^2-3x - 10 = 0`

`=> x = (3+-sqrt(9+4*10))/2 = 3/2+-7/2` (by the quadratic formula)

Normally we would be done, however originally we had `log(x-3)` in the equation, meaning we have the restriction `x>3`, as the logarithm function is undefined in the reals for `x<=0`.

As `3/2-7/2 < 3` it is not a solution.

As `3/2+7/2 = 5 > 3`, it is a solution.

Thus we have the solution `x = 5`