How do you evaluate #log_8 2#?

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Answer 1 · 2016-09-03T00:50:41

Start by remembering that the log function is the inverse of the exponential function.

#log_2 8 = y => 2^y = 8 #

2 is the base of the logarithm, and of the exponential also.

2 raised to what power equals 8?

#2^3 = 8#, right?

So, #log_2 8# equals 3.

Answer 2 · 2016-09-03T01:02:17

#1/3#

#log _"8"2 = x#

#8^x=2#

#x=1/3#