Solving Log Equation Double Integration - e^(e+1)

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SUMMARY

The discussion focuses on solving the log equation involving double integration of the functions y = LN x and y = e + 1 - x. The key finding is that x = e is a unique solution to the equation LN x = e + 1 - x, as LN x is strictly increasing while e + 1 - x is strictly decreasing. The participants suggest that graphical methods may be necessary for further exploration, but no elementary techniques for solving the equation are identified.

PREREQUISITES
  • Understanding of logarithmic functions, specifically LN x.
  • Familiarity with exponential functions and their properties.
  • Knowledge of double integration techniques in calculus.
  • Graphical interpretation of functions and their intersections.
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  • Study the properties of logarithmic and exponential functions in detail.
  • Learn techniques for double integration, particularly in bounded areas.
  • Explore graphical methods for solving equations involving transcendental functions.
  • Investigate numerical methods for finding roots of equations like LN x = e + 1 - x.
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Students and educators in calculus, mathematicians dealing with transcendental equations, and anyone interested in advanced integration techniques.

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Homework Statement



Do anyone know how to solve:
y=LN x
y=e+1-x


Homework Equations




y=LN x
y=e+1-x



The Attempt at a Solution



LN x=e+1-x
x=e^e * e^1 / e^x
x*e^x=e^(e+1)

then I don't know how to solve, must I solve it graphically?

Actually, the original question is :
Double integration
InIn x dxdy for the area bounded by y=e+1-x and y=LNx and x-axis
 
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Trial and error shows [itex]x = e[/itex] is a solution to [itex]\ln{x} = e + 1 - x[/itex]. Closer inspection reveals the solution is unique because [itex]\ln{x}[/itex] is strictly increasing while [itex]e + 1 - x[/itex] is strictly decreasing.

I know not of an elementary technique for solving such an equation.
 
Thanks a lot.
 

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