How do you solve ${log}_{9} (2 x + 5) = {log}_{9} 5 x$ $\log _ { 9} ( 2x + 5) = \log _ { 9} 5x$ ?

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Answer 1 · 2017-03-02T19:04:32

$x = \frac{5}{3}$ $x=5/3$

Because you have a ${log}_{9}$ $log_9$ on both sides, raise both sides as powers of $9$ $9$ , which effectively cancels out the $log$ $log$ s.

${log}_{9} (2 x + 5) = {log}_{9} (5 x)$ $log_9(2x+5) = log_9(5x)$

$9^{{log}_{9} (2 x + 5)} = 9^{{log}_{9} (5 x)}$ $9^(log_9(2x+5)) = 9^(log_9(5x))$

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

Now you can simply solve

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

$5 = 3 x$ $5=3x$

$x = \frac{5}{3}$ $x=5/3$

How do you solve \log _ { 9} ( 2x + 5) = \log _ { 9} 5xlog9(2x+5)=log95x\log _ { 9} ( 2x + 5) = \log _ { 9} 5x?