How do you solve $2 {log}_{8} (2 x - 5) = 4$ $2log_8(2x-5)=4$ ?

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How do you solve $2 {log}_{8} (2 x - 5) = 4$ $2log_8(2x-5)=4$ ?

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Answer 1 · 2015-11-25T23:52:01

$x = \frac{69}{2}$ $x = 69/2$

Explanation:

First of all, your domain is $2 x - 5 > 0 \Leftrightarrow x > \frac{5}{2}$ $2x - 5 > 0 <=> x > 5/2$ 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 $2$ $2$ :

$\times 2 {log}_{8} (2 x - 5) = 4$ $color(white)(xx)2 log_8(2x-5) = 4$
$\Leftrightarrow {log}_{8} (2 x - 5) = 2$ $<=> log_8(2x-5) = 2$

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

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

$\Leftrightarrow 8^{{log}_{8} (2 x - 5)} = 8^{2}$ $<=> 8^(log_8(2x-5)) = 8^2$

$\Leftrightarrow 2 x - 5 = 64$ $<=> 2x - 5 = 64$

$\Leftrightarrow 2 x = 69$ $<=> 2x = 69$

$\Leftrightarrow x = \frac{69}{2}$ $<=> x = 69/2$

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How do you solve $2 {log}_{8} (2 x - 5) = 4$ $2log_8(2x-5)=4$ ?

1 Answer

Explanation:

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