How do you solve ${log}_{2} (4 x - 8) = 1$ $log _ 2 (4x-8)=1$ ?

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How do you solve ${log}_{2} (4 x - 8) = 1$ $log _ 2 (4x-8)=1$ ?

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Answer 1 · 2016-01-31T10:05:48

$x = \frac{5}{2}$ $x = 5/2$

Explanation:

using : ${log}_{b} a = n \Leftrightarrow a = b^{n}$ $log_b a = n hArr a = b^n$

then ${log}_{2} (4 x - 8) = 1 \to 4 x - 8 = 2^{1} = 2$ $log_2 (4x - 8 ) = 1 → 4x - 8 = 2^1 = 2$

so 4x - 8 = 2 $\to 4 x = 10 \to x = \frac{10}{4} = \frac{5}{2}$ $→ 4x = 10 → x = 10/4 = 5/2$

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How do you solve ${log}_{2} (4 x - 8) = 1$ $log _ 2 (4x-8)=1$ ?

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How do you solve log2(4x−8)=1log _ 2 (4x-8)=1?

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Explanation:

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How do you solve ${log}_{2} (4 x - 8) = 1$ $log _ 2 (4x-8)=1$ ?