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

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In summary, the equation e^x+x=5 can be solved using the Lambert W-function or a numerical solution. It cannot be solved using only algebraic methods. The curve of e^x and 5-x intersect at one point, indicating one solution. An iterative method can be used to approximate the root of the equation, with a starting value of x=1 resulting in a solution of x=1.3066.
  • #1
physicsdreams
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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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  • #2


Nope, can't be solved with algebraic methods. It's Lambert W or a numerical solution.
 
  • #3


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

What is ln(e^x)?
 
  • #4


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.
 
  • #5


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
 

1. What is the first step in solving e^x+x=5 with Algebra?

The first step is to isolate the term with the exponent, e^x, by subtracting x from both sides of the equation.

2. Can e^x be simplified further?

No, e^x is already in its simplest form and cannot be simplified any further.

3. What is the next step after isolating e^x in the equation e^x+x=5?

The next step is to take the natural logarithm of both sides of the equation. This will eliminate the exponent and allow you to solve for x.

4. How can you solve for x if e^x is not isolated?

If e^x is not isolated, you can still solve for x by using the substitution method. Let u = e^x and substitute it into the equation to solve for x.

5. Are there any other methods to solve e^x+x=5 besides using Algebra?

Yes, you can also solve this equation using numerical methods, such as the Newton-Raphson method or the bisection method. These methods involve approximating the solution rather than solving it algebraically.

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