http://www.mathhelpforum.com/math-he...b08b1c96-1.gif

So we know that http://www.mathhelpforum.com/math-he...cdcb7831-1.gif

and that http://www.mathhelpforum.com/math-he...453d96a6-1.gif

Now solve those simultaneously

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- Jul 14th 2008, 07:23 AMracersteve47How would you go about solving simultaneously?
http://www.mathhelpforum.com/math-he...b08b1c96-1.gif

So we know that http://www.mathhelpforum.com/math-he...cdcb7831-1.gif

and that http://www.mathhelpforum.com/math-he...453d96a6-1.gif

Now solve those simultaneously - Jul 14th 2008, 07:56 AMflyingsquirrel
Hello

$\displaystyle

\begin{cases}

85=C\exp(3b) & (1)\\

13=C\exp(5b) & (2)

\end{cases}$

As $\displaystyle \exp(5b)\neq 0$, one can divide (1) by (2) which will give an equation that can be solved for $\displaystyle b$. Then use either (1) or (2) to find the value of $\displaystyle C$.