How do you evaluate #log_sqrt8 4096#?

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Answer 1 · 2016-11-17T03:06:22

#log_(sqrt8)4096=8#

Let #log_(sqrt8)4096=x#

Using definition of logarithm that #log_ab=m# means #a^m=b#

#(sqrt8)^x=4096#

expressing them in powers of #x#

or #(sqrt2^3)^x=2^12#

or #((2^(1/2))^3)^x=2^12#

or #2^((3x)/2)=2^12#

or #(3x)/2=12#

i.e. #x=12xx2/3=8#

Hence #log_(sqrt8)4096=8#