How do you solve $2+log_3(2x+5)-log_3x=4$ ?

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Answer 1 · 2015-02-08T01:29:13

I would start by collecting the logs on one side:

$log_3(2x+5)-log_3(x)=4-2$

I can use the fact that:
$logM+logN=log(M/N)$
Giving:

$log_3((2x+5)/x)=2$

Use the definition of logarithm:

$log_ax=b ->a^b=x$

Giving:

$(2x+5)/x=3^2$
$2x+5=9x$
$7x=5$
$x=5/7$

hope it helps

How do you solve 2+log_3(2x+5)-log_3x=42+log_3(2x+5)-log_3x=4?