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Alright, I'm helping a friend, but can't seem to be able to crack this question :

If
$\log_3 20 = a$,
$\log_3 15 = b$

then how do we represent with a,b
$\log_2 360$?

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2  
Try factoring 360. Remember that $\log(ab)=\log\,a+\log\,b$ –  J. M. Sep 8 '11 at 19:24
    
Is the base of the final logarithm $3$ or $2$? –  Srivatsan Sep 8 '11 at 19:27
    
i'm sure its 2. –  Bartlomiej Lewandowski Sep 8 '11 at 19:29
2  
Uh, @Craig, regarding homework: if s/he hasn't said that what s/he has is homework, it isn't our business to be tagging the question that way. –  J. M. Sep 8 '11 at 19:30

2 Answers 2

up vote 2 down vote accepted

First, we realize that $\log_3 5 = b-1$. Then $\log_3 4 = a-(b-1)$, and $\log_3 2 = (a-(b-1))/2$. So $\log_2 3 = 2 / (a-(b-1))$ and also $\log_2 5 = \log_3 5 / \log_3 2 = 2(b-1)/(a-(b-1))$. Finally, $\log_2 360 = 3 + \log_2 45 = 3 + 2\log_2 3 + \log_2 5 = 3 + \frac{2b+2}{a-b+1}$.

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A variant of the solution by @Craig is to observe that $$360=20\times 2\times 9=2\cdot 3^{a+2}\qquad\text{(Equation $1$)}$$ But we have also $$360=15\times 8\times 3=8\cdot 3^{b+1} \qquad\text{(Equation $2$)}$$

From the first equation we obtain $$\log_2 360=1+(a+2)\log_2 3 \qquad\text{(Equation $3$)}$$

From the second equation we obtain $$\log_2 360=3+(b+1)\log_2 3 \qquad\text{(Equation $4$)}$$

Use Equations $3$ and $4$ to solve for $\log_2 3$ in terms of $a$ and $b$, and substitute in Equation $3$.

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