Question

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

Answer 1

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`