Question 1162871
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^{2x}\,-\,26\,\cdot\,5^x\ +\ 25\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^{2x}\,-\,26\,\cdot\,5^x\ =\ -25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^x\(5^2\,-\,26\)\ =\ -25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^x(-1)\ =\ -25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^x\ =\ 25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^x\ =\ 5^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_5\(5^x\)\ =\ \log_5\(5^2\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\log_5(5)\ =\ 2\log_5(5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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