Question 1082041
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Your symbols and your text say two different things.

First you wrote "logarithmic equation" and then you wrote 9^-r = 9^1-2r


The symbols indicate an exponential equation, not logarithmic.


Then you said that -r and 1-2r are subscripts.  If that were true then your equation would be


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9_{-r}\ =\ 9_{1-2r}]


Which makes no sense at all without some additional context.


subscript -- sub, below, like a submarine


superscript -- super, above, like superman


So here is what I think you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9^{-r}\ =\ 9^{1\,-\,2r}]


Since *[tex \Large a^m\ =\ a^n] if and only if *[tex \Large m\ =\ n] we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -r\ =\ 1\ -\ 2r]


Solve for *[tex \Large r]


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">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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