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.

So where did all these answers come from and how to solve them. Please help me out! Thanks everyone!

  • 2
    $\begingroup$ Wellcome to MSE! It would be nice to use proper formatting. $\endgroup$
    – Wuestenfux
    Commented Jun 3, 2019 at 9:11
  • $\begingroup$ Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site. $\endgroup$ Commented Jun 3, 2019 at 9:27
  • 1
    $\begingroup$ I don't understand what the expression in the first question means. What is the meaning of $9\sqrt{3}$ ? $\endgroup$
    – Matti P.
    Commented Jun 3, 2019 at 9:28
  • $\begingroup$ @Wuestenfux got it! thanks $\endgroup$
    – Avallie
    Commented Jun 3, 2019 at 9:50
  • $\begingroup$ @N.F.Taussig gotcha thanks $\endgroup$
    – Avallie
    Commented Jun 3, 2019 at 9:50

1 Answer 1


Hint: $9\sqrt{3} = 3^2 \cdot 3^{1/2}= 3^{5/2}$

$\frac{1}{81}\sqrt{3} = 3^{-4} \cdot 3^{1/2} = 3^{-7/2}$

$\log_{a}{b} = c \implies a^c = b$


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