How do you solve $7^{2 x} = 2$ $7^(2x)=2$ ?

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How do you solve $7^{2 x} = 2$ $7^(2x)=2$ ?

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Answer 1 · 2015-10-18T11:35:03

$x = \frac{{log}_{7} (2)}{2}$ $x=log_7(2)/2$ .

Explanation:

We need to isolate the variable. Since the logarithm is the inverse function of the power, which means that ${log}_{a} a^{x} = x$ $log_a a^x = x$ , if we take the logarithm base $7$ $7$ of both members we get

${log}_{7} (7^{2 x}) = {log}_{7} (2)$ $log_7 (7^{2x}) = log_7(2)$ .

For what we have just observed, it becomes

$2 x = {log}_{7} (2) \Rightarrow x = \frac{{log}_{7} (2)}{2}$ $2x=log_7(2) \implies x=log_7(2)/2$ .

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How do you solve $7^{2 x} = 2$ $7^(2x)=2$ ?

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