How Do You Solve (1+x)e^-x = 0.5 for x?

  • Thread starter Xeract
  • Start date
In summary: Just trying to help.In summary, the problem involves multiplying both sides of an equation by a negative number. It can be solved using the Lambert W function.
  • #1
Xeract
5
0
Sorry if this is a very simple question, I am trying to rearrange (1+x)e^-x = 0.5 for x, and just can't seem to get my head around it. Any tips would be greatly appreciated.
 
Physics news on Phys.org
  • #2
Take log on both sides and go from there.
 
  • #3
Take the natural log of both sides and then use the properties of log to split the left side.

[tex]\ln(a/b)=\ln a - \ln b[/tex]

And remember that [tex]\ln e = 1 = \mbox{cancels each other out}[/tex]
 
Last edited:
  • #4
This is where I've got to, thanks for the responses:

ln((1+x)/e^x) = ln0.5
ln(1+x)-x =ln0.5
 
Last edited:
  • #5
I can't seem to go anywhere from your last step, is the problem written down correctly?
 
  • #6
do you have to solve this analytically?
 
  • #7
It's certainly impossible to solve analytically.
 
  • #8
nicksauce said:
It's certainly impossible to solve analytically.

yes, that's why I asked the OP ;)
 
  • #9
In general, a problem involving "x" both in an exponent and not can only be solved using the "Lambert W function" which is defined as the inverse function to f(x)= xex- and may require manipulation of the equation to put it in that form.
 
  • #10
I don't know if I'm doing this right, but here are my thoughts. if you factor out the first side you get xe^x+e^x=0.5 Then if you divide both sides by x you get e^x+e^x=0.5/x add your like terms on the left and you get 2e^x=0.5/x (i think that's right)
See if you can get it from there
((i just realized that this was from over a week ago AFTER i finished lol))
 
  • #11
llemes4011 said:
I don't know if I'm doing this right, but here are my thoughts. if you factor out the first side you get xe^x+e^x=0.5 Then if you divide both sides by x you get e^x+e^x=0.5/x add your like terms on the left and you get 2e^x=0.5/x (i think that's right)
See if you can get it from there
((i just realized that this was from over a week ago AFTER i finished lol))
That's wrong, read all the posts before yours.
 
  • #12
sorry, i also didn't read the original question right *sigh*
 
  • #13
llemes4011 said:
sorry, i also didn't read the original question right *sigh*
It's ok, no worries.
 

FAQ: How Do You Solve (1+x)e^-x = 0.5 for x?

What is the equation "Rearranging (1+x)e^-x = 0.5" used for?

The equation "Rearranging (1+x)e^-x = 0.5" is used for finding the value of x that satisfies the equation, or in other words, the value of x that makes the equation true.

What is the meaning of the "e" in the equation?

The "e" in the equation represents the mathematical constant approximately equal to 2.718, known as Euler's number. It is commonly used in equations involving growth and decay, such as in this case.

How do I solve this equation?

To solve this equation, you can use algebraic manipulation techniques to isolate the variable, in this case x, on one side of the equation. This involves multiplying, dividing, and taking logarithms of both sides until you have x by itself. You can also use numerical methods, such as graphing or using a calculator, to approximate the solution.

Why is the equation equal to 0.5?

The equation is equal to 0.5 because it represents the point at which the function (1+x)e^-x intersects the horizontal line y=0.5. In other words, it is the x-value where the function has a value of 0.5.

What are some real-life applications of this equation?

This equation can be used in various scenarios involving growth and decay, such as in population dynamics, radioactive decay, and bacterial growth. It can also be used in economics and finance to model exponential growth or decline of investments or loans.

Back
Top