How do you solve $log_2 (x+2) - log_2 (x-5) = 3$ ?

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Answer 1 · 2016-01-02T15:48:10

Unify the logarithms and cancel them out with $log_(2)2^3$

$x=6$

$log_(2)(x+2)+log_(2)(x-5)=3$

Property $loga-logb=log(a/b)$

$log_(2)((x+2)/(x-5))=3$

Property $a=log_(b)a^b$

$log_(2)((x+2)/(x-5))=log_(2)2^3$

Since $log_x$ is a 1-1 function for $x>0$ and $x!=1$ , the logarithms can be ruled out:

$(x+2)/(x-5)=2^3$

$(x+2)/(x-5)=8$

$x+2=8(x-5)$

$x+2=8x-8*5$

$7x=42$

$x=42/7$

$x=6$

How do you solve log_2 (x+2) - log_2 (x-5) = 3log_2 (x+2) - log_2 (x-5) = 3?