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October 25th, 2015, 12:45 PM   #1
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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
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October 25th, 2015, 03:18 PM   #2
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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.
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October 26th, 2015, 04:29 PM   #3
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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 $\displaystyle \sqrt[3]{\dfrac12}<1$
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October 28th, 2015, 07:58 AM   #4
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Quote:
Originally Posted by greg1313 View Post
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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