Solving the system of equations $12.3=D_0 e^{-Q/60R}$ and $24.1=D_0 e^{-Q/360R}$

Question

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

I have the following system equations:$$\begin{cases}12.3=D_0 e^{-Q/60R}\\ 24.1=D_0 e^{-Q/360R}\end{cases}$$

I have to solve for $D_0$, 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 $\frac{D_0}{D_0}$ which leaves me with no variable to solve for.

Any help would be greatly appreciated.

Thanks

Edit

For all those who want to see how to do it: \begin{align*} \frac{12.3}{24.1} & = \frac{D_0}{D_0} \frac{e^{-Q/60R}}{e^{-Q/360R}}\\ \frac{12.3}{24.1} & = e^{-Q/60R -(-Q/360R)}\\ \frac{12.3}{24.1} & = e^{-Q(1/60R + 1/360R)}\\ \ln \frac{12.3}{24.1} & = -Q(1/60R+ 1/360R)\\ Q & = -\frac{\ln \frac{12.3}{24.1}}{1/60R + 1/360R} \end{align*}

Answer 1

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 $D_0$.

Answer 2

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:

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

Answer 3

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.