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April 19th, 2014, 09:50 AM  #1 
Newbie Joined: Apr 2014 From: South Florida Posts: 8 Thanks: 1  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. 
April 19th, 2014, 11:58 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, scomora! We will use the identity: $\displaystyle \:\log_ba \:=\:\frac{\log_c a}{\log_c b}$ Quote:
$\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: $\displaystyle \;\frac{\log x}{\log y} \:=\:\frac{9}{2}$  
April 19th, 2014, 12:00 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
1. $\displaystyle \log_3\sqrt{243\sqrt{81\sqrt[3]{3}}}=\log_3\left(3^{\frac52}\cdot3\cdot3^{\frac{1 }{12}}\right)=\frac52+1+\frac{1}{12}=\frac{43}{12}$ $\displaystyle \log_2\sqrt[4]{64}10=\log_22^{\frac32}10=\frac3210=\frac{17}{2}$ $\displaystyle \frac{43/12}{17/2}=\frac{43}{102}$ 2. $\displaystyle \log_x64=\log_y\sqrt[3]{16}$ $\displaystyle \log_x2^6=\log_y2^{\frac43}$ $\displaystyle \frac{6\log2}{\log x}=\frac{\frac43\log2}{\log y}$ $\displaystyle \frac{\log x}{\log y}=\frac92$ Last edited by greg1313; April 20th, 2014 at 06:11 AM. 
April 19th, 2014, 06:42 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,281 Thanks: 1965 
I think you meant 17/2, not 17/10.


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