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October 25th, 2015, 12:45 PM  #1 
Newbie Joined: Oct 2015 From: Europe Posts: 1 Thanks: 0  I need help with logarithms http://akphoto3.ask.fm/828/331/746/.../preview/2.png http://akphoto3.ask.fm/319/584/697/.../preview/3.png can you show me how to do this step by step? I have other exercises then 
October 25th, 2015, 03:18 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond 
698: $\displaystyle 3\log_5(x4)>\dfrac{6}{\log_5(x4)+1}$ $\displaystyle 3\log_5^2(x4)+3\log_5(x4)6>0$ This is a quadratic in $\displaystyle \log_5(x4)$ with roots $\displaystyle 2,1$. Solve $\displaystyle x4=\dfrac{1}{25},x4=5$ and $\displaystyle \log_5(x4)+1=0$ to determine the values of interest for $\displaystyle x$. 709: Use the technique outlined above. Last edited by greg1313; October 25th, 2015 at 04:26 PM. 
October 26th, 2015, 04:29 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond 
711: $\displaystyle \log_{\sqrt[3]{\frac12}}\left(1\dfrac1x\right)\log_{\sqrt[3]{\frac12}}(x+1)<\log_{\sqrt[3]{\frac12}}\dfrac1x+3$ Use $\displaystyle \log a\log b=\log\dfrac ab$ and $\displaystyle \log a+\log b=\log a\cdot b$ to establish the inequality $\displaystyle \dfrac{x1}{x(x+1)}>\dfrac{1}{2x}$. Note that $\displaystyle 3=\log_{\sqrt[3]{\frac12}}\dfrac12$ and that $\displaystyle \sqrt[3]{\dfrac12}<1$ 
October 28th, 2015, 07:58 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
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