What is the solution to log base 4 (64) + 15*log base 32 (2)?

[SomeRandomGuy]

compute the following:

log base 4 (64) + 15*log base 32 (2)

I can't remember how to solve something like this. Makes me feel like an idiot, but I figured it never hurts to ask.

 

[Integral]

You need the basic definition of a log.

C = log_B A iff B^C = A

Apply that definition and it should be pretty easy.

 

[dextercioby]

compute the following:

log base 4 (64) + 15*log base 32 (2)

I can't remember how to solve something like this. Makes me feel like an idiot, but I figured it never hurts to ask.

I have strong reasons to believe that your answer is 6.
To prove that for yourself,try to apply the following formula
\log_{a}b =\frac{\log_{c} b}{\log_{c} a}
twice.

Daniel.

PS.I assumed you're familiar with this one:
\log_{c} x^{y} = y\log_{c} x.

 

[HallsofIvy]

Integral's way is much simpler than dextercioby's.

64= 4³ so log₄ 64= ?

32= 2⁵ so 2= 32^1/5. What is log₃₂ 2?