How do you solve #7^(2x)=2#?

1 Answer
KillerBunny
Oct 18, 2015

#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#, if we take the logarithm base #7# of both members we get

#log_7 (7^{2x}) = log_7(2)#.

For what we have just observed, it becomes

#2x=log_7(2) \implies x=log_7(2)/2#.

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