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$ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.

Any hints? Thanks!

colormegone
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Ovi
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  • A start might be to use the change-of-base formula to write both sides in terms of base-n logarithms. – colormegone May 28 '13 at 02:15
  • I tried that among many other things, I'll try again I guess – Ovi May 28 '13 at 02:15
  • I'll withdraw that first suggestion -- it's messy. Use the definition of logarithm: if both sides of the equation are equal to $ \ p \ $ , then $ \ \frac{(4n)^p}{(3n)^p} \ = \ (\frac{4}{3})^p \ $ is equal to what? The ratio will tell you the value of $ \ p \ $ and you can find $ \ n \ $ (or $ \ n^3 \ $ ) in turn. – colormegone May 28 '13 at 02:33
  • Yeah its equal to 8sqrt(3)/9 but it get's really really messy when solving for n and cubing it, and you would probably need a calculator unless you wanna spend an hour simplifying it. Isn't there a simpler way? – Ovi May 28 '13 at 02:39
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    You don't need a calculator: $ \ \frac{8 \sqrt{3}}{9} \ = \ \frac{8}{3 \sqrt{3}} \ $ . – colormegone May 28 '13 at 02:40
  • Yes but that's just the value of (4/3)^p. If you actually try to solve for n^3 it gets very messy. You could probably simplify it, but it would take a while. – Ovi May 28 '13 at 02:44

3 Answers3

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Here's a kind of ugly way of doing it: $$\begin{align*} (12n)^{\log_{4n}(40\sqrt{3})}&=(12n)^{\log_{3n}(45)}\\\\ (3)^{\log_{4n}(40\sqrt{3})}(4n)^{\log_{4n}(40\sqrt{3})}&=(4)^{\log_{3n}(45)}(3n)^{\log_{3n}(45)}\\\\ (3)^{\log_{4n}(40\sqrt{3})}\cdot 40\sqrt{3}&=(4)^{\log_{3n}(45)}\cdot 45\\\\ (3)^{\log_{4n}(40\sqrt{3})+\frac{1}{2}-2}&=(4)^{\log_{3n}(45)-\frac{3}{2}}\\\\ (3)^{\log_{4n}(40\sqrt{3})+\frac{1}{2}-2}&=(3)^{\log_3(4)\cdot(\log_{3n}(45)-\frac{3}{2})}\\\\ \log_{4n}(40\sqrt{3})-\frac{3}{2}&=\log_3(4)\cdot\left(\log_{3n}(45)-\frac{3}{2}\right)\\\\ \log_{3n}(45)-\frac{3}{2}&=\log_3(4)\cdot\left(\log_{3n}(45)-\frac{3}{2}\right)\\\\ \end{align*}$$ Therefore $$\begin{align*} \log_{3n}(45)&=\frac{3}{2}\\\\ 2025&=(3n)^3\\\\ n^3&=\frac{2025}{27}=75 \end{align*}$$

Zev Chonoles
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3

My proposal was that

$$\frac{(4n)^p}{(3n)^p} \ = \ (\frac{4}{3})^p \ = \ \frac{40 \sqrt{3}}{45} \ = \ \frac{8}{3 \sqrt{3}} \ = \ (\frac{4}{3})^{3/2} \ . $$

So both logarithms equal 3/2 and the rest follows as Zev Chonoles shows.

EDIT: Just to verify that the other piece checks,

$$(40 \sqrt{3})^2 \ = \ 4800 \ = \ (4n)^3 \ \Rightarrow \ n^3 \ = \ \frac{4800}{64} \ = 75 \ . $$

colormegone
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By elementary arithmetic operations (after / describing next action): $$\log_{4n}40\sqrt{3}=\log_{3n}45\ \ \ \mbox{ / definition of logarithm}$$ $$(4n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } 4=\frac{4}{3}\cdot 3$$ $$\left({4\over 3}\cdot 3n\right)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / }(ab)^c=a^cb^c$$ $$\left({4\over 3}\right)^{\log_{3n}45}\cdot (3n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } a^{\log_ab}=b$$ $$\left({4\over 3}\right)^{\log_{3n}45}\cdot 45=40\sqrt{3}\ \ \ \mbox { / }\cdot\frac{1}{45}$$ $$\left({4\over 3}\right)^{\log_{3n}45}={8\over 9}\sqrt{3}\ \ \ \mbox { / }\left({a\over b}\right)^c=\frac{a^c}{b^c}$$ $$\left({4\over 3}\right)^{\log_{3n}45}=\left({4\over 3}\right)^{3\over 2}\ \ \ \mbox { / }a^b=a^c\Rightarrow b=c\ \ (a\neq 0,1)$$ $$\log_{3n}45=\frac{3}{2}\ \ \ \mbox { / definition of logarithm}$$ $$(3n)^{3\over 2}=45\ \ \ \mbox { / powered by } \frac{2}{3}\mbox{ and divided by }3$$ $$n=\frac{\sqrt[3]{45^2}}{3}\ \ \ \mbox { / powered by 3}$$ $$n^3=\frac{45^2}{27}$$ $$n^3=75$$

Bartek Pawlik
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