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May 29th, 2014, 06:46 AM   #1
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log(x ،3 ) +log(25 ،3)+ log(9 ، x)= 3+ log (25 ، x)
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May 29th, 2014, 07:10 AM   #2
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$\displaystyle \log { \left( 3\cdot x \right) } +\log { \left( 25\cdot 3 \right) +\log { \left( 9\cdot x \right) =3+\log { \left( 25\cdot x \right) } } } \\ \\ \\ \log { \left( 3x \right) +\log { \left( 75 \right) } } +\log { \left( 9x \right) } =\log { \left( 1000 \right) +\log { \left( 25x \right) } } \\ \\ \\ \log { \left( 25x \right) -\log { \left( 3x \right) } } -\log { \left( 9x \right) } =\log { \left( 75 \right) } -\log { \left( 1000 \right) } \\ \\ \\ \log { \left( 25x \right) } -\left[ \log { \left( 3x \right) +\log { \left( 9x \right) } } \right] =\log { \left( \frac { 75 }{ 1000 } \right) } \\ \\ \log { \left( 25x \right) } -\log { \left( 27{ x }^{ 2 } \right) } =\log { \left( \frac { 75 }{ 1000 } \right) } \\ \\ \log { \left( \frac { 25x }{ 27x\cdot x } \right) } =\log { \left( \frac { 75 }{ 1000 } \right) } \\ \\ \log { \left( \frac { 25 }{ 27x } \right) =\log { \left( \frac { 75 }{ 1000 } \right) } } \\ \\ \frac { 25 }{ 27x } =\frac { 75 }{ 1000 } \\ \\ \frac { 25000 }{ 27x } =75\\ \\ 2025x=25000\\ \\ x=\frac { 25000 }{ 2025 } =\frac { 1000 }{ 81 } \\ \\ $
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May 29th, 2014, 07:30 AM   #3
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thanks for you

but imean
log(x،3) =logx/log3
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May 29th, 2014, 07:35 AM   #4
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apologies, didn't know what that symbol represented.
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May 29th, 2014, 07:54 AM   #5
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Do you need an exact answer, or can it be in decimals?

So you need a solution for this:

$\displaystyle \frac { \log _{ 10 }{ x } }{ \log _{ 10 }{ 3 } } +\frac { \log _{ 10 }{ 25 } }{ \log _{ 10 }{ 3 } } +\frac { \log _{ 10 }{ 9 } }{ \log _{ 10 }{ x } } =\frac { \log _{ 10 }{ 27 } }{ \log _{ 10 }{ 3 } } +\frac { \log _{ 10 }{ 25 } }{ \log _{ 10 }{ x } } $

Last edited by perfect_world; May 29th, 2014 at 08:01 AM.
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May 29th, 2014, 08:23 AM   #6
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yes it is thanks for you perfect_world
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May 29th, 2014, 08:27 AM   #7
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$\displaystyle \frac { \log _{ 10 }{ x } }{ \log _{ 10 }{ 3 } } +\frac { \log _{ 10 }{ 25 } }{ \log _{ 10 }{ 3 } } +\frac { \log _{ 10 }{ 9 } }{ \log _{ 10 }{ x } } =\frac { \log _{ 10 }{ 27 } }{ \log _{ 10 }{ 3 } } +\frac { \log _{ 10 }{ 25 } }{ \log _{ 10 }{ x } } \\ \\ \frac { \log _{ 10 }{ x } }{ \log _{ 10 }{ 3 } } +\frac { \log _{ 10 }{ 25 } }{ \log _{ 10 }{ 3 } } -\frac { \log _{ 10 }{ 27 } }{ \log _{ 10 }{ 3 } } =\frac { \log _{ 10 }{ 25 } }{ \log _{ 10 }{ x } } -\frac { \log _{ 10 }{ 9 } }{ \log _{ 10 }{ x } } \\ \\ \frac { \log _{ 10 }{ \left( \frac { 25x }{ 27 } \right) } }{ \log _{ 10 }{ 3 } } =\frac { \log _{ 10 }{ \left( \frac { 25 }{ 9 } \right) } }{ \log _{ 10 }{ x } } \\ \\ \log _{ 3 }{ \left( \frac { 25x }{ 27 } \right) } =\log _{ x }{ \left( \frac { 25 }{ 9 } \right) } \\ \\ { 3 }^{ \log _{ x }{ \left( \frac { 25 }{ 9 } \right) } }=\frac { 25x }{ 27 } \\ \\ { 3 }^{ 3 }\cdot { 3 }^{ \log _{ x }{ \left( \frac { 25 }{ 9 } \right) } }=25x\\ \\ { 3 }^{ 3+\log _{ x }{ \left( \frac { 25 }{ 9 } \right) } }=25x\\ \\ x=\frac { { 3 }^{ 3+\log _{ x }{ \left( \frac { 25 }{ 9 } \right) } } }{ 25 } \\ \\ Use:\\ \\ \\ { x }_{ n+1 }=\frac { { 3 }^{ 3+\log _{ { x }_{ n } }{ \left( \frac { 25 }{ 9 } \right) } } }{ 25 } \\ \\ { x }_{ 0 }=2\\ \\ \therefore \quad { x }_{ 1 }=\quad 5.453\quad \left( 3dp \right) \\ \\ { x }_{ 2 }=2.093\quad \left( 3dp \right) \\ \\ { x }_{ 3 }=4.936\quad \left( 3dp \right) \\ \\ ...\quad { x }_{ 300+ }=3,\quad \therefore \quad x=3\\ \\ $

That's only one solution, but there seem to be 2 according to this graph: https://www.desmos.com/calculator/ex5e6rteor

Last edited by perfect_world; May 29th, 2014 at 08:40 AM.
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May 30th, 2014, 01:40 AM   #8
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Math Focus: Physics, mathematical modelling, numerical and computational solutions
This equation has analytical results

$\displaystyle \frac{\log x}{\log 3} + \frac{\log 25}{\log 3} + \frac{\log 9}{\log x} = 3 + \frac{\log 25}{\log x}$
$\displaystyle \frac{(\log x)^2}{\log 3} + \frac{\log 25 \cdot \log x}{\log 3} + \log 9 = 3 \log x + \log 25$
$\displaystyle \frac{(\log x)^2}{\log 3} + \left(\frac{\log 25}{\log 3} - 3\right)\log x + \log 9 - \log 25 = 0$

let $\displaystyle y = \log x$,

$\displaystyle \frac{y^2}{\log 3} + \left(\frac{\log 25}{\log 3} - 3\right)y + \log 9 - \log 25 = 0$

Solutions of quadratic are determined using the formula

$\displaystyle y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

where $\displaystyle a = \frac{1}{\log 3}$, $\displaystyle b = \left(\frac{\log 25}{\log 3} - 3\right)$ and $\displaystyle c = \log 9 - \log 25$.

$\displaystyle y = \frac{3 - \frac{\log 25}{\log 3} \pm \sqrt{\left(\frac{\log 25}{\log 3} - 3\right)^2 - \frac{4\log \frac{9}{25}}{\log 3}}}{\frac{2}{\log 3}}$
$\displaystyle = \frac{3\log 3 - \log 25 \pm \sqrt{\left(\log 25 - 3\log 3\right)^2 - 4\log \frac{9}{25} \cdot \log 3}}{2}$
$\displaystyle = \frac{3\log 3 - \log 25 \pm \sqrt{(\log 25)^2 - 6\log 25 \cdot \log 3+ 9(\log 3)^2 - 4\log \frac{9}{25} \cdot \log 3}}{2}$
$\displaystyle = \frac{3\log 3 - \log 25 \pm \sqrt{(\log 25)^2 - 6\log 25 \cdot \log 3+ 9(\log 3)^2 - 4(2\log 3 - \log 25) \log 3}}{2}$
$\displaystyle = \frac{3\log 3 - \log 25 \pm \sqrt{(\log 25)^2 - 6\log 25 \cdot \log 3+ 9(\log 3)^2 - 8(\log 3)^2 + 4\log 25 \cdot \log 3}}{2}$
$\displaystyle = \frac{3\log 3 - \log 25 \pm \sqrt{(\log 25)^2 - 2\log 25 \cdot \log 3+ (\log 3)^2}}{2}$
$\displaystyle = \frac{3\log 3 - \log 25 \pm \sqrt{(\log 25 - \log 3)^2}}{2}$
$\displaystyle = \frac{3\log 3 - \log 25 \pm (\log 25 - \log 3)}{2}$

Therefore,

$\displaystyle y = \frac{3\log 3 - \log 25 + (\log 25 - \log 3)}{2}
$
$\displaystyle y = \frac{2\log 3}{2} = \log 3$
$\displaystyle x = 10^y = 10^{\log 3} = 3$

$\displaystyle y = \frac{3\log 3 - \log 25 - (\log 25 - \log 3)}{2} = 2\log 3 - \log 25 = \log\frac{9}{25}
$
$\displaystyle x = 10^y = 10^{\log(\frac{9}{25})} = \frac{9}{25}$
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