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Say I have the following two equations:

$$2000 = k \ln {\frac{t+1}{t}}$$ $$2000 = k \ln {\frac{t+3}{t+1}}$$

The solution for $t$ is $t=1$, which is easily obtained equating the two equations and exponentiating both sides.

In Maple, this is:

e1 := 2000 = k*ln((t + 1)/t)
e2 := 2000 = k*ln((t + 3)/(t + 1))
solve({ e1, e2 }, { t })

The $solve$ command produces no result above. If instead I do

e3 := k*ln((t + 1)/t) = k*ln((t + 3)/(t + 1))
solve(e3, t)

I obtain the correct answer $t=1$.

Why doesn't the solve command work when $e1$ and $e2$ are considered a system of two equations?

xoux
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  • In the first piece of code, why is t in curly braces? – mjw Aug 19 '21 at 02:30
  • In the first system you are explicity setting e1 and e2 to $2000$, maybe that has something to do with it? – mjw Aug 19 '21 at 02:32
  • The curly braces are because that is the syntax I saw in the book I am following, "Understanding Maple" by Ian Thompson. – xoux Aug 19 '21 at 21:43
  • Maple programming questions such as this (which are not primarily about mathematics) are off-topic for this forum. More appropriate are forums www.mapleprimes.com or stackoverflow.com – acer Aug 22 '21 at 13:06

1 Answers1

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e1 := 2000 = k*ln((t + 1)/t)
e2 := 2000 = k*ln((t + 3)/(t + 1))
solve({ e1, e2 }, { t , k})

works fine, and returns the solution. You asked for maple to solve the system for t but to keep k arbitrary -- there is indeed no solution then.

Jacques Carette
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