How do you solve ${log}_{6} x + {log}_{6} (x + 1) = 1$ $log_6x + log_6(x+1) =1$ ?

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Answer 1 · 2015-07-29T16:09:01

$x = 2$ $x=2$

The left side has two logs with same base x, hence these two can be combined to give ${log}_{6} x (x + 1) = 1$ $log_6 x(x+1) =1$

This implies $x (x + 1) = 6$ $x(x+1)=6$ ,

$x^{2} + x - 6$ $x^2 +x-6$ =0
$(x + 3) (x - 2) = 0$ $(x+3)(x-2)=0$
$x = - 3$ $x=-3" "$ , $x = 2$ $x=2$

But $x = - 3$ $x=-3$ is an extraneous solution, because ${log}_{6} (- 3)$ $log_6(-3)$ is not a real number.

How do you solve log_6x + log_6(x+1) =1log6x+log6(x+1)=1log_6x + log_6(x+1) =1?