Question #e7110

1 Answer
Nathan L.
Aug 16, 2017

#x ~~ 0.047#

Explanation:

We're asked to solve for #x# in

#16/15 = 4^x#

We can change this into logarithmic form:

#log_4(16/15) = x#

#x = log_4(16/15) = (log(16/15))/(log(4)) ~~ color(blue)(ulbar(|stackrel(" ")(" "0.047" ")|)#

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