evaluate log base 6 of the fifth root of 36

Question

whpalmer4 · Answer

Hint: $$\sqrt[5]{36} = 36^{\frac{1}{5}}$$

anonymous · Answer

okay, but I'm kind of stuck because 36 is the square of 6 and I don't know how I should evaluate it with the 1/5
if that makes any sense at all

whpalmer4 · Answer

okay, if 6 is the square root of 36, then $36 = 6^2$ and we can rewrite our problem as
$$\sqrt[5]{36} = (6^2)^{\frac{1}{5}}$$Now, do you remember how to simplify $$(a^b)^c$$?

anonymous · Answer

um
that would be a^b x c, right?

whpalmer4 · Answer

$$(a^b)^c = a^{b*c}$$
so that gives us
$$6^{\frac{2}{5}}$$How would we find 
$$\log_{6}6^{\frac{2}{5}}$$

anonymous · Answer

oh my gosh
thank you so much
that makes a lot more sense

whpalmer4 · Answer

hint: $$\log_{b}b^n=n$$

anonymous · Answer

okay, so the solution would be 2/5
thank you very much

whpalmer4 · Answer

exactly!

whpalmer4 · Answer

one little quibble from earlier:
you wrote "that would be a^b x c, right?"
I know what you meant, but really you should have written that as a^(b x c) because the operator precedence is that exponentiation comes before multiplication, so what you actually wrote is c x a^b

whpalmer4 · Answer

just trying to avoid a potential wrong answer for you down the road :-)

anonymous · Answer

okay, okay :)
thanks for getting what i mean, despite my bad form