How do you evaluate #log_2 (1/4)#?

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Answer 1 · 2016-08-17T17:49:01

To understand logs, read #log_2 (1/4)# as asking the question..

"What index of 2 will give #1/4#"?

Write #1/4# in a different form.# rarr " "1/4 = 1/2^2 = 2^(-2)#

"What index of 2 will give #2^(-2)#?

The answer is obviously -2.

#:. log_2 (1/4) = -2#

Or convert #log_2 (1/4)# to index form.

#log_2 (1/4) = x " "rArr 2^x = 1/4#

Solve the exponential equation by making the bases the same.

#2^x = 1/4 = 1/2^2 = 2^-2#

If # 2^x = 2^-2 #

#x = -2#