How do you solve $2log_8(2x-5)=4$ ?

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Answer 1 · 2015-11-25T23:52:01

$x = 69/2$

First of all, your domain is $2x - 5 > 0 <=> x > 5/2$ since the argument of any logarithmic expression needs to be greater than zero.

Now, to solve the equation, you should first divide both sides of the equation by $2$ :

$color(white)(xx)2 log_8(2x-5) = 4$
$<=> log_8(2x-5) = 2$

The inverse function of $log_8(x)$ is $8^x$ . This means that $log_8(8^x) = x$ and $8^(log_8 x) = x$ .

In other words, you can make $log_8$ disappear by applying the exponential function $8^x$ on both sides!

$<=> 8^(log_8(2x-5)) = 8^2$

$<=> 2x - 5 = 64$

$<=> 2x = 69$

$<=> x = 69/2$

How do you solve 2log_8(2x-5)=42log_8(2x-5)=4?