How do you write the equation #log_8 (4)=2/3# in exponential form?

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Answer 1 · 2017-01-04T15:55:20

#4=8^(2/3)#

If #log _b a = k#, its inverse is #a=b^k#.

Here, the inverse of #log_8 4=2/3#, the inverse is #4=8^(2/3)#.

Some more examples:

RELATION INVERSE (for the same relation)

#y=ln x to x=e^y#

#y=log x to x=10^y#

#2=log_16 256 to 256=16^2#