Is there a way of doing these?

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- Nov 3rd 2009, 04:10 AMStuck ManSimultaneous equations with x and y exponents
Is there a way of doing these?

- Nov 3rd 2009, 04:36 AMJameson
That depends.

It's a good idea to ask for specific examples rather than really general questions. It's hard to answer statements like the one you wrote here.

The answer to your question depends on if you mean by hand vs. computer, only basic algebra versus logarithms, etc. - Nov 3rd 2009, 06:57 AMStuck Man
e^x + 3e^y = 3

e^2x - 9e^2y = 6

The answer has to be expressed as logs to base e.

I have the answers and they seem to be correct. - Nov 3rd 2009, 07:09 AMalexmahone
$\displaystyle e^x+3e^y=3$

$\displaystyle e^{2x}-9e^{2y}=6$

$\displaystyle (e^x)^2-(3e^y)^2=6$

$\displaystyle (e^x+3e^y)(e^x-3e^y)=6$

$\displaystyle 3(e^x-3e^y)=6$

$\displaystyle e^x-3e^y=2$

$\displaystyle 2e^x=5$

$\displaystyle e^x=\frac{5}{2}$

$\displaystyle x=ln (\frac{5}{2})$

$\displaystyle 6e^y=1$

$\displaystyle e^y=\frac{1}{6}$

$\displaystyle y=ln(\frac{1}{6})=-ln 6$ - Nov 4th 2009, 03:42 AMStuck Man
Thanks.