How do you solve # log_6 (x+ 1)+ log_6 (x-4)= 1 #?

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Answer 1 · 2016-07-30T22:39:11

#x=5#

Given:

#log_6(x+1)+log_6(x-4) = 1#

We require:

#6 = 6^1 = 6^(log_6(x+1)+log_6(x-4)) = (x+1)(x-4) = x^2-3x-4#

Subtract #6# from both ends to get:

#0 = x^2-3x-10 = (x-5)(x+2)#

So #x=5# or #x=-2#

If #x=5# then:

#log_6(x+1)+log_6(x-4) = log_6(6)+log_6(1) = 1+0 = 1#

So #x=5# is a solution of the original equation.

If #x=-2# then (even if we allow Complex logarithms):

#log_6(x+1)+log_6(x-4)#

#= log_6(-1)+log_6(-6)#

#= log_6(1)+(pi i)/ln(6) + log_6(6)+(pi i)/ln(6)#

#= 1+(2pi i)/ln(6) != 1#

So #x=-2# is not a solution of the original equation.