How do you solve #4^ { 2x } = 11#?

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Answer 1 · 2017-05-17T14:31:52

Refer to explanation

To do this question, you have to use logs:

#2xlog4 = log11#

Then divide by log4 to isolate #x#:

#2x = (log11)/(log4)#

#2x = 1.73#

Divide by 2 to get #x# on its own:

#x = (1.73)/2#
#x= 0.87#