How do you solve $log(x-3)+log(x+5)=log9$ ?

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Answer 1 · 2017-01-04T01:55:36

$x=4$ , and possibly $x=-6$ , depending upon the context of the question and what led to the logarithmic equarion

$log(x-3)+log(x+5)=log9$

Using $logAB=logA+logB$ we get:

$log(x-3)(x+5)=log9$

Using $log A=log B <=> A=B$ we have

$(x-3)(x+5)=9$
$:. x^2+5x-3x-15=9$
$:. x^2+2x-24=0$
$:. (x+6)(x-4)=0$
$:. x=4,-6$

Now depending upon the context of the question we could probably eliminate $x=-6$ .

$x=-6$ , in the original form of the question, would lead to the log of negative numbers which do not exist.

How do you solve log(x-3)+log(x+5)=log9log(x-3)+log(x+5)=log9?