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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: 7,958 Thanks: 1146 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: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 711: $\displaystyle \log_{\sqrt{\frac12}}\left(1-\dfrac1x\right)-\log_{\sqrt{\frac12}}(x+1)<\log_{\sqrt{\frac12}}\dfrac1x+3$ 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 $\displaystyle 3=\log_{\sqrt{\frac12}}\dfrac12$ and that $\displaystyle \sqrt{\dfrac12}<1$ 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. Tags logarithms Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zendetax Algebra 1 September 23rd, 2015 06:37 AM Shawn Algebra 4 May 23rd, 2013 11:50 AM kvrajiv Algebra 1 June 4th, 2012 10:46 AM jwood Algebra 3 March 25th, 2011 01:41 PM empiricus Algebra 8 July 9th, 2010 08:31 AM

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