Given #p=2^x# and #q = 2^y#, what is #log_4(4p^2)/q# in terms of #x and y#?
1 Answer
Aug 28, 2017
The expression equals
Explanation:
Substitute.
#=log_4((4(2^x)^2)/(2^y))#
We now use the properties of logarithms to simplify.
#=log_4 4 + log_4(2^x)^2 - log_4 2^y#
#= 1 + log_4(4^x) - log_4 2^y#
#= 1 + xlog_4 4 - ylog_4 2#
#= 1 + x - 1/2y#
Hopefully this helps!
