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I am busy with 1st year engineering and I am struggling with this seemingly trivial math problem.

I have to two equations:

$\ 12.3=Dₒe^{-Q/60R}$


$\ 24.1=Dₒe^{-Q/360R}$

I have to solve for Dₒ where Q is an unknown but constant in both equations and R = 8.314 (Gas Constant). My lecturer was saying something about dividing the one equation by the other but that results in Dₒ/Dₒ which leaves me with no variable to solve for.

Any help would be greatly appreciated.


EDIT - For all those who want to see how to do it

$\frac{12.3}{24.1}$= $\frac{Dₒ}{Dₒ}$ $\frac{e^{-Q/60R}}{e^{-Q360R}}$

$\frac{12.3}{24.1}$= $\ e^{-Q/60R -(-Q/360R)}$

$\frac{12.3}{24.1}$= $\ e^{-Q(1/60R + 1/360R)}$

$\ ln{12.3/24.1}$ = $\ -Q(1/60R+ 1/360R)$

$\ Q=$ $\ -(ln{12.3/24.1}) / (1/60R + 1/360R)$

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$\log$ both sides to get a more familiar linear system. – wj32 May 8 '13 at 21:04
Thank you guys for all the help! I was also forgetting my basic exponent rules. Not the best thing to forget the night before semester tests. – Keagan Ladds May 8 '13 at 21:13
Should I not rather ln both sides? – Keagan Ladds May 8 '13 at 21:22
Yes, $\ln$ is another way of writing $\log$. – wj32 May 9 '13 at 0:50
up vote 1 down vote accepted

Dividing one equation by the other will give $\frac{12.3}{21.1} = \exp(-Q(\frac{1}{60R} - \frac{1}{360R}))$, which you can then solve for $Q$. Plugging $Q$ in to one of the equations will then let you solve for $D_0$.

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If $Q$ is a constant, then dividing one equation by the other will eliminate $D_{0}$ and allow you to solve for $Q$. Then once you have determined $Q$, plug into one of your equations to find, for example:

$D_{0} = \frac{12.3}{e^{-Q/60R}}$

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Hint: let $$x=e^{-Q/360R}$$

Note that $x^6 = e^{-Q/60R}$.

Solve for $x$ and $D_0$, then work out $Q$ from there.

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