How do you evaluate #log_16(1/4)#?

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Answer 1 · 2016-03-23T14:01:34

#x=-1/2#

Let #log_16(1/4)=x#, then #16^x=1/4# or

#(2^4)^x=1/(2^2)# or

#2^(4x)=2^(-2)#

Hence #4x=-2# i.e. #x=-2/4=-1/2#

Answer 2 · 2016-03-23T14:10:13

Let's write it in exponential form.

#log_ab = x -> a^x = b#

#log_16(1/4) = x -> 16^x = 1/4#

Solve for x by putting everything in the same base.

#(2^4)^x = 1/(2^2)#

#2^(4x) = 2^(-2)#

#x = -1/2#

Therefore, #log_16(1/4) = -1/2#

Practice exercises:

a). #log_9(1/27)#

b) #log_x(81)= 4#

c) #log_(2x + 1)11x = 2#

Good luck!