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

Related questions
See all questions in Functions with Base b
Impact of this question
1938 views around the world
You can reuse this answer
Creative Commons License