How do you solve ${log}_{2} (6 x + 9) - {log}_{2} (3) = 3$ $log_2(6x+9)-log_2(3)=3$ ?

Question

1690 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-24T05:33:41

$x = \frac{5}{2}$ $x=5/2$

Use the following rule: ${log}_{a} (b) - {log}_{a} (c) = {log}_{a} (\frac{b}{c})$ $log_a (b)-log_a (c)=log_a (b/c)$

${log}_{2} (6 x + 9) - {log}_{2} (3) = 3$ $log_2(6x+9)-log_2(3)=3$

$\Rightarrow {log}_{2} (\frac{6 x + 9}{3}) = 3$ $=>log_2((6x+9)/3)=3$

$\Rightarrow {log}_{2} (2 x + 3) = 3$ $=>log_2(2x+3)=3$

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

$2^{{log}_{2} (2 x + 3)} = 2^{3}$ $2^(log_2(2x+3))=2^3$

$2 x + 3 = 8$ $2x+3=8$

$2 x = 5$ $2x=5$

$x = \frac{5}{2}$ $x=5/2$

How do you solve log2(6x+9)−log2(3)=3log_2(6x+9)-log_2(3)=3?