Efficient Variable Solving with Maple: Logarithm Properties and Equations

[uman]

Homework Statement

$${\frac {65-75\,{e^{-5\,k}}}{1-{e^{-5\,k}}}}={\frac {60-75\,{e^{-10\,k} }}{1-{e^{-10\,k}}}}$$

Homework Equations

$${e}^{x}=k$$ implies $$x=\ln \left( k \right)$$, as well as other properties of logarithms.

The Attempt at a Solution

Maple makes short work of this, giving $$k=1/5\,\ln \left( 2 \right)$$, but I'm totally lost as to how to solve it myself.

 

[Rainbow Child]

I you set
$$e^{-5\,k}=z$$
then what
$$e^{-10\,k}$$
equals to?

 

[Architeuthis Dux]

As a scientist, it is important to understand and be able to solve equations efficiently and accurately. The use of Maple and other software can certainly make this process easier, but it is also important to have a solid understanding of the underlying concepts and principles involved. In this case, the use of logarithm properties can greatly simplify the equation and lead to a quick solution.

The given equation involves exponential functions, which can be solved using logarithmic functions. The first step would be to isolate the exponential terms on one side of the equation by multiplying both sides by the denominators. This would result in:

65-75e^(-5k) = 60-75e^(-10k)

Next, we can use the property that e^x = k implies x = ln(k) to rewrite the exponential terms as logarithms. This would give us:

65-75ln(1/e^(5k)) = 60-75ln(1/e^(10k))

We can simplify further by using the property that ln(1/x) = -ln(x). This would give us:

65+75ln(e^(5k)) = 60+75ln(e^(10k))

Now, we can use the property that ln(e^x) = x to simplify further:

65+75(5k) = 60+75(10k)

Simplifying, we get:

325k = 600k

Finally, we can solve for k by dividing both sides by 325, giving us:

k = 1/5 ln(2)

This is the same solution that Maple provided, but by understanding the properties of logarithms and using them to simplify the equation, we are able to solve it ourselves. This not only helps to build a deeper understanding of the concept, but also allows us to solve similar equations in the future without relying on software.