1. ## Thinking/InquiryProblem Solving: Logarithms

I am having some difficulty with this question. All I really know how to do is rewrite it.

If $\displaystyle \log{a}{b} = \frac{1}{x}$ and $\displaystyle \log{b}{\sqrt{a}} = 3x^2$, show that $\displaystyle x = \frac{1}{6}$

*note: I don't know how to do those logs better... it is saying "log base a" and the other one is "log base b"*

I have tried everything I could think of. I subbed in 1/6 for x, made them into exponent form, but I don't know how to answer this question. Or even what it's asking.

Help?

2. $\displaystyle \log_b\sqrt{a}=\frac{1}{2}\log_ba=3x^2\Rightarrow\ log_ba=6x^2$.

But $\displaystyle \log_ab\cdot\log_ba=1$.

Then $\displaystyle \frac{1}{x}\cdot 6x^2=1\Rightarrow 6x=1\Rightarrow x=\frac{1}{6}$