How do you simplify #log_4 8#?

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Answer 1 · 2016-05-07T16:10:59

Use the logarithmic properties:

#log_a(b^c)=c*log_a(b)#

#log_a(b)=log_c(b)/log_c(a)#

You can notice that #c=2# fits this case since #8# can be derived as a power of #2#. Answer is:

#log_(4)8=1.5#

#log_(4)8#

#log_(2)8/log_(2)4#

#log_(2)2^3/log_(2)2^2#

#(3*log_(2)2)/(2*log_(2)2)#

#3/2#

#1.5#