How do you solve $log_5 (x+3)= 3 + log_5 (x-3)$ ?

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Answer 1 · 2016-04-10T09:51:47

$x=189/62=3 3/62$

As $log_ba=loga/logb$

$log_5(x+3)=3+log_5(x-3)$ can be written as

$log(x+3)/log5=3+log(x-3)/log5$

As $log5!=0$ , multiplying both sides by $log5$

$log(x+3)=3log5+log(x-3)$ or

$(x+3)=5^3(x-3)=125(x-3)$ or

$x+3=125x-375$ or

$124x=378$ and $x=378/124=189/62=3 3/62$

How do you solve log_5 (x+3)= 3 + log_5 (x-3) log_5 (x+3)= 3 + log_5 (x-3) ?