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 September 21st, 2013, 02:52 PM #1 Newbie   Joined: Sep 2013 Posts: 2 Thanks: 0 Exponential Hi guys, need help with... (18-5^x)/5^x =5^x+2
 September 21st, 2013, 03:01 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Exponential $\frac{18\,-\,5^x}{5^x}\,=\,5^x\,+\,2 \\ 18\,-\,5^x\,=\,5^{2x}\,+\,2\,\cdot\,5^x \\ 5^{2x}\,+\,3\,\cdot\,5^x\,-\,18\,=\,0 \\ $$5^x\,+\,6$$$$5^x\,-\,3$$\,=\,0 \\ 5^x\,=\,3\,\Rightarrow\,x\,=\,\frac{\ln3}{\ln5}$
 September 21st, 2013, 03:07 PM #3 Newbie   Joined: Sep 2013 Posts: 2 Thanks: 0 Re: Exponential Why did you write 5^2x and 3? Where did they come from?
 September 21st, 2013, 05:15 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: Exponential I multiplied both sides of your equation by 5^x and then rearranged the terms. 5^x + 2 * 5^x = 3 * 5^x. When multiplying exponentials with the same base, add the exponents: 5^x * 5^x = 5^(x + x) = 5^(2x).

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