How do you solve $log_2(6x+9)-log_2(3)=3$ ?

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Answer 1 · 2015-12-24T05:33:41

$x=5/2$

Use the following rule: $log_a (b)-log_a (c)=log_a (b/c)$

$log_2(6x+9)-log_2(3)=3$

$=>log_2((6x+9)/3)=3$

$=>log_2(2x+3)=3$

Exponentiate both sides with base $2$ to undo the logarithm.

$2^(log_2(2x+3))=2^3$

$2x+3=8$

$2x=5$

$x=5/2$

How do you solve log_2(6x+9)-log_2(3)=3log_2(6x+9)-log_2(3)=3?