System of equations.

Can you tell us the context of the problem? I suspect it's at least a contest-type problem; I'll admit I used technology to speed my thinking.

The second equation can be expressed in terms of k2/k1 or k1/k2, and you can show that there is only one solution; but the only way I can see to solve it is by inspection (or by graphing). Find that constant, and you can express k2 in terms of k1.

The first equation, if you stare at it long enough, amounts to finding two points on the graph of ln(x) such that the slope between them is 828. You can soon see that x and y must both be very small.

But now replace k2 with your expression in terms of k1 and see what you get.

Here is a graph of the two equations, zoomed in appropriately:

FMH120054.png
 
Here is a hint. It is hard to work with these unwieldy expressions. Try the following substitutions and reduce both equations to what n equals. Note that k_1 and k_2 obviously cannot be equal. How does that help you?

[MATH]m = \dfrac{k_2}{(k_2 - k_1)} \text { and } n = ln \left ( \dfrac{k_2}{k_1} \right ).[/MATH]
 
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JeffM said:
Here is a hint. It is hard to work with these unwieldy expressions. Try the following substitutions and reduce both equations to what n equals. Note that k_1 and k_2 obviously cannot be equal. How does that help you?

[MATH]m = \dfrac{k_2}{(k_2 - k_1)} \text { and } n = ln \left ( \dfrac{k_2}{k_1} \right ).[/MATH]
Click to expand...
By the way, the method I indicated will, if you are careful, lead to a purely algebraic solution.
 
The second equation can be rewritten as [MATH] \frac{\text{k2} \log \left(\frac{2 \text{k1}}{\text{k2}}\right)}{\text{k2}-\text{k1}}=0 [/MATH], so [MATH]k1=\frac{k2}{2}[/MATH]. Substituting in the first equation, we have [MATH]k2=\frac{\log2}{414}[/MATH].
 
Tavasanis said:
… [MATH]k1=\frac{k2}{2}[/MATH] … [MATH]k2=\frac{\log2}{414}[/MATH]
Click to expand...
Hi Tavasanis. I'm thinking maybe you switched your subscripts while typing.

Using the equations as given, I went through the substitution method and got ln(2)/414 for k1 -- and half that for k2. In other words, my result (below) matches yours with the subscripts switched:

k2 = k1/2 … k1 = ln(2)/414

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