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

1 Answer
May 3, 2018

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!