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 September 14th, 2011, 06:53 AM #1 Newbie   Joined: Sep 2011 Posts: 2 Thanks: 0 Algebra and Power Help... Please! Hello all, I need help solving the following equation for x: a = (1-x) ^ {[2/(1+g)]-2} Can anyone help? Thank you so much.
 September 14th, 2011, 07:11 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Algebra and Power Help... Please! Hello ejfl, $a= (1-x) ^ {\{[2/(1+g)]-2\}}$ $\log(a)=\log((1-x) ^ {\{[2/(1+g)]-2\}}$ $\log(a)=\{[2/(1+g)]-2\} \cdot \log(1-x)$ $\log(1-x)=\frac{\{[2/(1+g)]-2\}}{\log(a)}$ Can you proceed from here?
 September 14th, 2011, 09:06 AM #3 Newbie   Joined: Sep 2011 Posts: 2 Thanks: 0 Re: Algebra and Power Help... Please! Thanks for the prompt response Hoempa - can I just confirm that the denominator and numerator on the right are the correct way round?
 September 14th, 2011, 09:44 AM #4 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Algebra and Power Help... Please! Well, I slipped up $\log(a)=\{[2/(1+g)]-2\} \cdot \log(1-x) \\ \\ \\ \frac{\log(a)}{\{[2/(1+g)]-2\}} = \frac{\{[2/(1+g)]-2\} \cdot \log(1-x)}{\{[2/(1+g)]-2\}} = \log(1-x)$ $\{[2/(1+g)]-2\} \ne 0$
 September 15th, 2011, 05:27 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Algebra and Power Help... Please! WARNING! Mistakes in this post! (Can you find them?) $a\,=\,(1\,-\,x)^{\frac{2}{1+g}-2}$ $\log_n(a)\,=\,$$\frac{-2g}{1+g}$$\log_n(1\,-\,x)$ $-\frac{(1+g)\log_n(a)}{2g}\,=\,\log_n(1\,-\,x)$ $a^{$$-(1+g)\log_n(a)/2g$$}\,=\,1\,-\,x$ $x\,=\,1\,-\,a^{$$-(1+g)\log_n(a)/2g$$}$
 September 15th, 2011, 10:00 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,976 Thanks: 2228 If a = 1/4 and g = 1/2, what does your result tell you x is?
 September 15th, 2011, 04:55 PM #7 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Algebra and Power Help... Please! $-\frac{(1+g)\log_n(a)}{2g}\,=\log_n(a) \cdot \left(-\frac{1+g}{2g}\right) =\,\log_n(1\,-\,x)$ $n^{\log_n(a) \cdot \left(-\frac{1+g}{2g}\right) }= n^{\log_n(1\,-\,x)}$ $a^{\left(-\frac{1+g}{2g}\right)}= 1-x$ $1-a^{\left(-\frac{1+g}{2g}\right)}= x$
 September 16th, 2011, 01:52 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,976 Thanks: 2228 You could get your last line from greg1313's first line.
September 16th, 2011, 02:08 AM   #9
Math Team

Joined: Apr 2010

Posts: 2,780
Thanks: 361

Re: Algebra and Power Help... Please!

Thank you I was biased by logs..

$a\,=\,(1\,-\,x)^{\frac{2}{1+g}-2} = (1\,-\,x)^{\frac{-2g}{1+g}}$

So (with z=x^y yields x=y^(1/z)

$1-x= a^{-\frac{1+g}{2g}}$

$x=1-a^{-\frac{1+g}{2g}}$

Quote:
 Originally Posted by greg1313 $-\frac{(1+g)\log_n(a)}{2g}\,=\,\log_n(1\,-\,x)$ $a^{$$-(1+g)\log_n(a)/2g$$}\,=\,1\,-\,x$
after your last edit, these lines aren't equivalent. Generally,
$a^{n \log(b)} \ne b$; choose a = 2 and n = 4, b = 16 for example.

 September 16th, 2011, 02:24 AM #10 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Algebra and Power Help... Please! Its been a while since I've done logs... Thanks!

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