How do you solve #log_3 3 + log_2 8 + log_4 64#?

1 Answer
Gió
Nov 19, 2015

I found #7#.

Explanation:

You can use the definition of log as:
#log_bx=a -> x=b^a#
So, basically you can substitute the exponents of the base that you need to get the argument:
#log_3(3)=1# because #3^1=3#
#log_2(8)=3# because #2^3=8#
#log_4(64)=3# because #4^3=64#
And finally:
#1+3+3=7#

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