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May 1st, 2014, 01:26 AM   #1
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logarithm inequality

Log base 7(3x - 1) + log base 7 (2x + 1) < 0. I understand that (3x - 1) >0 and 2x + 1> 0. That gives me x > 1/3 and x > - 1/2 but the second inequality is wrong how?
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May 1st, 2014, 02:05 AM   #2
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$\displaystyle \color{blue}{x>\dfrac{1}{3}\Longrightarrow\ x \in (\dfrac{1}{3},\ \infty )\ \qquad (1) \\\;\\ AND \\\;\\x>-\dfrac{1}{2}\Longrightarrow\ x \in (-\dfrac{1}{2},\ \infty)\ (2)\\\;\\(1), \ (2)\ \Longrightarrow\ x \in (\dfrac{1}{3},\ \infty ) \cap (-\dfrac{1}{2},\ \infty) \Longrightarrow\ x \in (\dfrac{1}{3},\ \infty )}$

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May 1st, 2014, 05:48 AM   #3
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The second inequality is not correct according to my answers
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May 1st, 2014, 05:57 AM   #4
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Quote:
Originally Posted by bongantedd View Post
The second inequality is not correct according to my answers
?!
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May 1st, 2014, 05:58 AM   #5
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The second inequality is not correct according to my answers. I got the same answer as yours. Does that mean we are wrong or my book is wrong. The answer is 1/2<x >1/3 bt how???
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May 1st, 2014, 06:22 AM   #6
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$\displaystyle (\dfrac{1}{3},\ \infty)$

it is the field of existence (of inequality).

Inequality solution (after some calculations ! )

is the interval

$\displaystyle (\dfrac{1}{3},\ \dfrac{1}{2})\\\;\\ So,\qquad \dfrac{1}{3} <x < \dfrac{1}{2}$
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May 1st, 2014, 06:56 AM   #7
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$\displaystyle \color{blue}{Conditions\ for\ logarithms :\\\;\\3x-1 > 0 \Rightarrow x>\dfrac{1}{3}\Longrightarrow\ x \in (\dfrac{1}{3},\ \infty )\ \qquad (1) \\\;\\ 2x+1 > 0 \Rightarrow x>-\dfrac{1}{2}\Longrightarrow\ x \in (-\dfrac{1}{2},\ \infty)\ (2)\\\;\\(1), \ (2)\ \Longrightarrow\ x \in (\dfrac{1}{3},\ \infty ) \cap (-\dfrac{1}{2},\ \infty) \Longrightarrow\ x \in (\dfrac{1}{3},\ \infty )\ \ (3)\\\;\\ \log_7(3x-1)+\log_7(2x+1) < 0\ \Rightarrow\ \log_7(3x-1)(2x+1) < 0 \\\;\\(3x-1)(2x+1) < 7^0 \Rightarrow 6x^2+3x-2x-1 < 1 \\\;\\6x^2+x-2 < 0 \Rightarrow x \in (-\dfrac{2}{3},\ \dfrac{1}{2})\ \ (4)\\\;\\(3), (4) \Rightarrow x \in (\dfrac{1}{3},\ \infty) \cap (-\dfrac{2}{3},\ \dfrac{1}{2}) \Rightarrow x \in (\dfrac{1}{3},\ \dfrac{1}{2}) \Rightarrow \dfrac{1}{3} < x < \dfrac{1}{2}}$
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