How do you solve $log (x + 9) = log 4 x$ $log(x + 9) = log4x$ ?

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Answer 1 · 2015-11-18T11:31:44

$x = 3$ $x=3$

For the logs to be equal then what you are taking logs of are also equal.

Write as

$log (x + 9) = log (4 x) \to x + 9 = 4 x$ $log(x+9)=log(4x) -> x+9 = 4x$

$3 x = 9$ $3x=9$
$x = 3$ $x=3$

How do you solve log(x+9)=log4xlog(x + 9) = log4x ?