Question

How do you solve `log(x − 3) + log(x − 2) = log(2x + 24)`?

Answer 1

`x = 9`

Explanation:

`log(x-3) + log(x-2) = log(2x+24)`

`log((x-3)(x-2)) = log(2x+24)`

`(x-3)(x-2) = 2x + 24`

Simplify the left side:
`x^2 - 3x - 2x + 6 = 2x + 24`

`x^2 - 5x + 6 = 2x + 24`

Add `color(blue)(5x)` on both sides of the equation:
`x^2 - 5x + 6 quadcolor(blue)(+quad5x) = 2x + 24 quadcolor(blue)(+quad5x)`

`x^2 + 6= 7x + 24`

Move everything to the left side of the equation so that we can factor:
`x^2 + 6 - 7x - 24 = 0`

`x^2 - 7x - 18 = 0`

To factor this, we have to find two numbers that:

• Add up to `-7`
• Multiply up to `-18` from (`1 * -18`)#

So we have to find a pair of factors of `-18` that add up to `-7`.

We know that the factors of `-18` are:
`-18, -9, -6, -3, -2, -1, 1, 2, 3, 6, 9, 18`. The two factors that add up to `-7` are `-9 and 2`.

So the factored form becomes:
`(x-9)(x+2) = 0`

So
`x-9 = 0` and `x+2 = 0`

`x = 9quad` and `quadx = -2`

However, these are not our final solutions! We must check our work when solving log equations by plugging it back into the original equation, since you cannot have a log of `0` or anything negative.

Let's check our first solution, `x = 9`:
`log(9-3) + log(9-2) = log(2(9) + 24)`

`log6 + log7 = log(18 + 24)`

`log(6*7) = log42`

`log42 = log42`

This is true! That means that `x = 9`.

Now let's check `x = -2`:
`log(-2-3) + log(-2-2) = log(2(-2)+24)`

`log(-5) + log(-4) = log(-4 + 24)`

Oh no! We cannot take the log of a negative number, but `-5` and `-4` are negative! That means that `x = -2` is NOT a solution.

Finally, the answer is `x = 9`.

Hope this helps!