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

#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!