How do you solve #log_2 (4x)=5#?

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Answer 1 · 2016-07-11T13:12:09

8

in a variety of ways

for example

#log_2 (4x)=5#

#implies log_2 4 + log_2 x=5#
#implies 2 + log_2 x=5#
#implies log_2 x=3#
#implies x=2^3 = 8#

OR

#log_2 (4x)=5#
#2^(log_2 (4x))=2^5#
#4x=32#
#x = 8#

OR

simplest, maybe, taken straight from the idea that a logarithm is just an index

#log_2 (4x)=5#
#implies 4x=2^5#
#implies x=2^5* 2^(-2) = 2^3 = 8#