Can e^x+x=5 be solved without graphing using algebraic methods?

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The equation e^x + x = 5 cannot be solved using traditional algebraic methods and requires the Lambert W-function or numerical solutions. Attempts to rearrange the equation, such as using ln(5-x), do not yield a straightforward solution. Graphical analysis indicates that the curves of e^x and 5-x intersect at one point, confirming there is only one solution. An iterative method can be employed to approximate the solution, starting with an initial guess and refining it through repeated substitutions. Ultimately, the equation's complexity necessitates advanced mathematical tools or numerical approaches for resolution.
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Homework Statement



e^x+x=5

Homework Equations



Lambert W-function?

The Attempt at a Solution



I moved everything around and got: Ln(5-x)=5...
It doesn't really help.
I looked at wolframalpha, and it said I need the Lambert W-function (no clue what that is).

Can this equation be solved Without graphing it, only using algebraic methods?
 
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Nope, can't be solved with algebraic methods. It's Lambert W or a numerical solution.
 


You might want to check your rearrangement of the original equation.

What is ln(e^x)?
 


physicsdreams said:

Homework Statement



e^x+x=5

Homework Equations



Lambert W-function?

The Attempt at a Solution



I moved everything around and got: Ln(5-x)=5...
It doesn't really help.
I looked at wolframalpha, and it said I need the Lambert W-function (no clue what that is).

Can this equation be solved Without graphing it, only using algebraic methods?

What you want to do. IF you want to find exact solution then i think it is task of machines(wolfarmalpha) and too tough for me.
If you want to know the number of solutions this equations will have then. It is possible.
draw curve of e^x and 5-x.rough e^x curve is as a^x(a>0). and 5-x is straight line with slope -1. So these two curve will intersect each other at one point hence it will have one solution.
 


You can get the root of the equation with an iterative method for the desired accuracy. Rearrange the equation:

x=ln(5-x)

Choose an x0 <5, determine x1=ln(5-x0) as the next approximation. Substitute again, to get the next x and repeat the procedure: xk+1=ln(5-xk)

Starting with x=1, the following values are obtained: 1.386, 1.284, 1.312, 1.305, 1.307, 1.3064, 13066, 1.3066

ehild
 
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