Efficient Variable Solving with Maple: Logarithm Properties and Equations

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The discussion focuses on solving the equation involving exponential terms using logarithm properties. The equation presented is a ratio of expressions that simplifies to a solution for k using Maple, yielding k = (1/5) ln(2). A key step in manual solving involves substituting e^(-5k) with a variable z, prompting the question of how to express e^(-10k) in terms of z. The conversation highlights the challenge of solving such equations without computational tools. Understanding the relationship between the exponential and logarithmic forms is crucial for manual resolution.
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Homework Statement


{\frac {65-75\,{e^{-5\,k}}}{1-{e^{-5\,k}}}}={\frac {60-75\,{e^{-10\,k}<br /> }}{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.
 
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I you set
e^{-5\,k}=z
then what
e^{-10\,k}
equals to?
 
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