How do you solve #\log _ { 6} ( x + 7) + \log _ { 6} ( x + 2) = 2#?

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Answer 1 · 2016-11-14T00:45:42

Start by using the rule #log_a(n) + log_a(m) = log_a(n xx m)#.

#log_6((x+ 7)(x + 2)) = 2#

#log_6(x^2 + 7x+ 2x + 14) = 2#

#log_6(x^2 + 9x + 14) = 2#

#x^2 + 9x + 14 = 36#

#x^2 + 9x - 22 = 0#

#(x + 11)(x - 2) = 0#

#x= -11 and 2#

Hoever, #x = -11# is extraneus since it renders the original equation undefined. #x= 2# is the only actual solution.

Hopefully this helps!