Logarithms in Systems of Equations

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SUMMARY

The discussion centers on solving the system of equations defined by the functions y=e^{-x}+1 and y=2+lnx. The user determined the solution set by identifying the intersection point of the plots, specifically (0.62745018..., 1.5339409...). It was concluded that while some equations may have straightforward algebraic solutions, this particular system does not lend itself to a neat analytical solution. The complexity of logarithmic and exponential functions in this context necessitates numerical methods for finding solutions.

PREREQUISITES
  • Understanding of logarithmic functions, specifically natural logarithms (ln).
  • Familiarity with exponential functions and their properties.
  • Knowledge of numerical methods for solving equations, such as root-finding algorithms.
  • Basic graphing skills to visualize function intersections.
NEXT STEPS
  • Study numerical methods for solving nonlinear equations, such as the Newton-Raphson method.
  • Learn about graphing techniques for visualizing intersections of functions.
  • Explore the properties of exponential and logarithmic functions in greater depth.
  • Investigate specific cases of equations that have analytical solutions, such as x + e^x = 1.
USEFUL FOR

Students, educators, and mathematicians interested in solving complex systems of equations involving logarithmic and exponential functions, as well as those seeking to improve their numerical analysis skills.

Mandelbroth
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Homework Statement


Consider the system...

[itex]y=e^{-x}+1 \\ y=2+lnx[/itex]

The Attempt at a Solution


I don't think there is a "pretty way" (algebraic manipulation) to do this. I simply found the point where the plots intercepted, getting the solution set (0.62745018..., 1.5339409...).

Is there a better way to do this? For what it's worth, this was for a friend who claims that her math teacher is unreasonable, so I don't know if it does has a fully algebraic approach.

Edit: For clarity, both have a domain and codomain in the set of real numbers, so there isn't an imaginary part to worry about.
 
Last edited:
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Mandelbroth said:
Is there a better way to do this?
In general, there is not. Some equations have an analytic solution which is easy to find (x+e^x = 1 for example), but the general case does not have a nice solution.
 

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