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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: $\displaystyle 3\log_5(x-4)>\dfrac{6}{\log_5(x-4)+1}$ $\displaystyle 3\log_5^2(x-4)+3\log_5(x-4)-6>0$ This is a quadratic in $\displaystyle \log_5(x-4)$ with roots $\displaystyle -2,1$. Solve $\displaystyle x-4=\dfrac{1}{25},x-4=5$ and $\displaystyle \log_5(x-4)+1=0$ 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 $\displaystyle \dfrac{x-1}{x(x+1)}>\dfrac{1}{2x}$. Note that $\displaystyle 3=\log_{\sqrt[3]{\frac12}}\dfrac12$ and that $\displaystyle \sqrt[3]{\dfrac12}<1$
October 28th, 2015, 07:58 AM   #4
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Quote:
 Originally Posted by greg1313 698: $\displaystyle 3\log_5(x-4)>\dfrac{6}{\log_5(x-4)+1}$ $\displaystyle 3\log_5^2(x-4)+3\log_5(x-4)-6>0$
This is true as long as $\displaystyle log_5(x-4)+ 1> 0$ which is the same as saying $\displaystyle log_5(x- 4)> -1$ or that $\displaystyle x- 4> 5^{-1}= 1/5$, $\displaystyle x> 4+ 1/5= 21/5$.

Quote:
 This is a quadratic in $\displaystyle \log_5(x-4)$ with roots $\displaystyle -2,1$. Solve $\displaystyle x-4=\dfrac{1}{25},x-4=5$ and $\displaystyle \log_5(x-4)+1=0$ to determine the values of interest for $\displaystyle x$. 709: Use the technique outlined above.

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