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April 19th, 2014, 09:50 AM   #1
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Logarithms - Somewhat Difficult

I've been trying to figure out these two problems for about a week now. I got -29/51 for the first one, but I just want to make sure it's right. For the second one, I was trying to solve for x and substitute it in, but I'm having trouble with it. All help would be appreciated!

P.s. I didn't use the math tags because I'm not really sure how to type an nth root without typing a rational exponent.
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April 19th, 2014, 11:58 AM   #2
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Hello, scomora!

We will use the identity: $\displaystyle \:\log_ba \:=\:\frac{\log_c a}{\log_c b}$


Quote:
$\displaystyle 2.\;\;\text{ If }\log_x64 \,=\,\log_y\sqrt[3]{16}\;\;\color{red}{[1]}$
$\displaystyle \quad \text{ find }\frac{\log x}{\log y}$

$\displaystyle \log_x64 \:=\:\frac{\log_264}{\log_2x} \:=\:\frac{\log_2(2^6)}{\log_2x} \:=\:\frac{6\log_2(2)}{\log_2x} \:=\:\frac{6\cdot1}{\log_2x} \:=\:\frac{6}{\log_2x} $

$\displaystyle \log_y(16)^{\frac{1}{3}} \:=\:\frac{\log_2(2^4)^{\frac{1}{3}}} {\log_2y}\:=\:\frac{\log_2(2^{\frac{4}{3}})}{\log_ 2y} \:=\:\frac{\frac{4}{3}\log_22}{\log_2y} \:=\: \frac{\frac{4}{3}}{\log_2y}$

Substitute into $\displaystyle \color{red}{[1]}\!:\;\;\frac{6}{\log_2x} \:=\:\frac{\frac{4}{3}}{\log_2y} \quad\Rightarrow\quad \frac{6}{\frac{4}{3}} \:=\:\frac{\log_2x}{\log_2y}$


Therefore: $\displaystyle \;\frac{\log x}{\log y} \:=\:\frac{9}{2}$

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April 19th, 2014, 12:00 PM   #3
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1.

$\displaystyle \log_3\sqrt{243\sqrt{81\sqrt[3]{3}}}=\log_3\left(3^{\frac52}\cdot3\cdot3^{\frac{1 }{12}}\right)=\frac52+1+\frac{1}{12}=\frac{43}{12}$

$\displaystyle \log_2\sqrt[4]{64}-10=\log_22^{\frac32}-10=\frac32-10=-\frac{17}{2}$

$\displaystyle \frac{43/12}{-17/2}=-\frac{43}{102}$

2.

$\displaystyle \log_x64=\log_y\sqrt[3]{16}$

$\displaystyle \log_x2^6=\log_y2^{\frac43}$

$\displaystyle \frac{6\log2}{\log x}=\frac{\frac43\log2}{\log y}$

$\displaystyle \frac{\log x}{\log y}=\frac92$
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Last edited by greg1313; April 20th, 2014 at 06:11 AM.
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April 19th, 2014, 06:42 PM   #4
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I think you meant 17/2, not 17/10.
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