How do you solve $2 + {log}_{3} (2 x + 5) - {log}_{3} x = 4$ $2+log_3(2x+5)-log_3x=4$ ?

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Answer 1 · 2015-02-08T01:29:13

I would start by collecting the logs on one side:

${log}_{3} (2 x + 5) - {log}_{3} (x) = 4 - 2$ $log_3(2x+5)-log_3(x)=4-2$

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

${log}_{3} (\frac{2 x + 5}{x}) = 2$ $log_3((2x+5)/x)=2$

Use the definition of logarithm:

${log}_{a} x = b \to a^{b} = x$ $log_ax=b ->a^b=x$

Giving:

$\frac{2 x + 5}{x} = 3^{2}$ $(2x+5)/x=3^2$
$2 x + 5 = 9 x$ $2x+5=9x$
$7 x = 5$ $7x=5$
$x = \frac{5}{7}$ $x=5/7$

hope it helps

How do you solve 2+log_3(2x+5)-log_3x=42+log3(2x+5)−log3x=42+log_3(2x+5)-log_3x=4?