How do you solve $log _2 (x+2) - log _2 (x-5) = 3$ ?

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Answer 1 · 2016-05-20T07:27:21

6

Use, $log_bm-log_bn=log_b(m/n)$ and, if $log_b a=c$ then $a=b^c$

$log_2 (x+2)-log_2 (x-5)$

$=log_2((x+2)/(x-5))=3$ . Inversely,

(x+2)/(x-5)= 2^3=8#. Solving,.

x=6..

How do you solve log _2 (x+2) - log _2 (x-5) = 3log _2 (x+2) - log _2 (x-5) = 3?