Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you evaluate #Log_2 (64)=x#?

1 Answer

Feb 4, 2015

You can use the definition of logarithm:
#log_a(b)=x# #-># #a^x=b#

In your case you get:

#2^x=64# which can be written as:

#2^x=2^6# (because #2^6=64#)

and finally (same base, #2# and comparing the exponents):

#x=6#

Related questions

See all questions in Logarithm-- Inverse of an Exponential Function

Impact of this question

3920 views around the world

You can reuse this answer
Creative Commons License