How do you solve ${log 5}^{2 x} = 4$ $log5^(2x) = 4$ ?

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Answer 1 · 2015-11-04T17:39:17

The solution is $x = \frac{2}{log (5)}$ $x=2/log(5)$

A property of lohgarithms states that

$log (a^{b}) = b log (a)$ $log(a^b)=blog(a)$

In your case, $a = 5$ $a=5$ and $b = 2 x$ $b=2x$ , so the equality becomes

$log (5^{2 x}) = 2 x log (5)$ $log(5^{2x})=2xlog(5)$ .

At this point, it's very easy to isolate the $x$ $x$ , since $log (5)$ $log(5)$ is just a number, and we have

$log (5^{2 x}) = 4$ $log(5^{2x}) = 4$

$\Leftrightarrow$ $iff$

$2 x log (5) = 4$ $2xlog(5)=4$

$\Leftrightarrow$ $iff$

$x = \frac{4}{2 log (5)} = \frac{2}{log (5)}$ $x=4/(2log(5)) = 2/log(5)$

How do you solve log52x=4log5^(2x) = 4?