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 June 27th, 2010, 12:30 PM #1 Newbie   Joined: Jun 2010 Posts: 22 Thanks: 0 logarithm Can you help me with this set of equations?? $5^x-2^y= 1 \ \ \ \ \ \ \ \\ 5^{x-2}\cdot 2^{y-2} = 0.2 \ \ \ \ \ \$
 June 27th, 2010, 12:51 PM #2 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 Re: logarithm Let $a= 5^x$ and $b= 2^y$.
 June 27th, 2010, 03:52 PM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: logarithm And write 5^(x-2) as (5^x)^(5^-2) ... (or multiply both sides by 25 and by 4)
 June 27th, 2010, 08:49 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: logarithm $5^x\,-\,2^y= 1\,\Rightarrow\,5^x\,=\,2^y\,+\,1$ $5^{x-2}\,\cdot\,2^{y-2}\,=\,0.2\,\Rightarrow\,\frac{2^y\,+\,1}{25}\,\cd ot\,2^{y\,-\,2}\,=\,\frac{1}{5}$ $\frac{2^{2y\,-\,2}\,+\,2^{y\,-\,2}}{25}\,=\,\frac{1}{5}$ $2^{2y\,-\,2}\,+\,2^{y\,-\,2}\,-\,5\,=\,0$ $\frac{2^{2y}}{4}\,+\,\frac{2^y}{4}\,-\,5\,=\,0$ $2^{2y}\,+\,2^y\,-\,20\,=\,0$ $a\,=\,2^y$ $a^2\,+\,a\,-\,20\,=\,0$ $(a\,+\,5)(a\,-\,4)\,=\,0$ $2^y\,=\,4$ $y\,=\,\frac{\ln{4}}{\ln{2}}\,=\,2\,\Rightarrow\,y\ ,=\,2,\,x\,=\,1$

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