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.

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