How do you solve #Log_2 x^7=21#?

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Answer 1 · 2016-05-28T21:50:25

#x=8#

Using basic properties of logs we find:

#21 = log_2 x^7 = 7 log_2 x#

Dividing both ends by #7# we have:

#3 = log_2 x#

So:

#2^3 = 2^(log_2 x) = x#

That is:

#x = 2^3 = 8#