# Logarithms / Maths with a root base or to the power of root

Greetings StackExchange! I would like to ask this question: How do I solve Logarithms, or just simply "Advanced Maths" (aka where these Logarithm numbers came from)

Here's the question:

1. $$\log_{9\sqrt{3}}\frac{1}{81}\sqrt{3}$$ (Note: the base is $$9\sqrt{3}$$ so it's pretty much $$9\sqrt{3}^x = \frac{1}{81}\sqrt{3}$$

The answer is $$-\frac{7}{5}$$ and I know that if it's $$\log_9 \frac{1}{81}$$, the result would simply be $$-2$$, but now that there are roots; I'm quite confused.

1. $$4\sqrt{2}\log_{8\sqrt{2}} 3$$ (Note: $$8\sqrt{2}$$ is the base so it's $$8\sqrt{2}^x = 3^{4\sqrt{2}}$$)

The answer is $$3^\frac{5}{7}$$ but I don't even know how to solve this one, so it should be $$3^{4\sqrt{2}}$$? which means $$3^{\sqrt{32}}$$, and $$\sqrt{32} = 32^\frac{1}{2}$$. So it's not $$3^{{32}^{\frac12}}$$ right? I'm quite confused.

• Wellcome to MSE! It would be nice to use proper formatting. – Wuestenfux Jun 3 '19 at 9:11
• Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site. – N. F. Taussig Jun 3 '19 at 9:27
• I don't understand what the expression in the first question means. What is the meaning of $9\sqrt{3}$ ? – Matti P. Jun 3 '19 at 9:28
• @Wuestenfux got it! thanks – Felicia Natalia Rivera Jun 3 '19 at 9:50
• @N.F.Taussig gotcha thanks – Felicia Natalia Rivera Jun 3 '19 at 9:50

Hint: $$9\sqrt{3} = 3^2 \cdot 3^{1/2}= 3^{5/2}$$
$$\frac{1}{81}\sqrt{3} = 3^{-4} \cdot 3^{1/2} = 3^{-7/2}$$
$$\log_{a}{b} = c \implies a^c = b$$