How do you solve #log[(x + 3)(x - 8)] + log[(x + 3)/(x - 8)] = 2#?

By the rules of logarithms, #log(m) + log(n) = log(mn)#, so

#log[(x+3)(x-8)] + log[(x+3)/(x-8)] = log[((x+3)(x-8)(x+3))/(x-8)]#

We cancel the (x-8) on the numerator and the denominator

#log[(x+3)(x-8)] + log[(x+3)/(x-8)] = log[(x+3)^2]#

The problem told us that that equaled 2, so

#log[(x+3)^2] = 2#

We know that if #log_b(a) = c# then #a = b^c# so we have

#(x+3)^2 = 10^2#

We use 10 because if the log doesn't have a base, it's implicitly 10. (Also note that we couldn't have put that square to the front of the logarithm and cut it with the 2 because we'd lose a root if we did it)

Now we take the square root,

#x + 3 = +-10#

And solve these values
#x = 10 - 3 = 7#
#x = -10 - 3 = -13#

But before that, check for the values we know x can't be! Logarithms can't take null or negative arguments so we know #x != 3 and x!=8#

We know that #x-3# and #x-8# can only be negative if the other is negative too. so we study the signs

------(-3)++++++++++++ #(x-3)#
----------------------(+8)++ #(x-8)#

++++0---------------0+++ #(x-3)(x-8) or (x-3)/(x-8)#

We know we can't take values that are 0 or negative, so #x > 8 or x < -3# which means there's only one possible root, -13.