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 Algebra Pre-Algebra and Basic Algebra Math Forum

April 19th, 2014, 08:50 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.
Attached Images Screen Shot 2014-04-19 at 12.45.39 PM.jpg (11.5 KB, 5 views) April 19th, 2014, 10:58 AM   #2
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Hello, scomora!

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

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

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

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

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

Therefore: $\displaystyle \;\frac{\log x}{\log y} \:=\:\frac{9}{2}$ April 19th, 2014, 11:00 AM #3 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond 1. $\displaystyle \log_3\sqrt{243\sqrt{81\sqrt{3}}}=\log_3\left(3^{\frac52}\cdot3\cdot3^{\frac{1 }{12}}\right)=\frac52+1+\frac{1}{12}=\frac{43}{12}$ $\displaystyle \log_2\sqrt{64}-10=\log_22^{\frac32}-10=\frac32-10=-\frac{17}{2}$ $\displaystyle \frac{43/12}{-17/2}=-\frac{43}{102}$ 2. $\displaystyle \log_x64=\log_y\sqrt{16}$ $\displaystyle \log_x2^6=\log_y2^{\frac43}$ $\displaystyle \frac{6\log2}{\log x}=\frac{\frac43\log2}{\log y}$ $\displaystyle \frac{\log x}{\log y}=\frac92$ Thanks from scomora Last edited by greg1313; April 20th, 2014 at 05:11 AM. April 19th, 2014, 05:42 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 I think you meant 17/2, not 17/10. Thanks from greg1313 and scomora Tags algebra, difficult, logarithms, logs, precalculus ,

### scomora

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