solve: x*e^x=1?

Question

1)its not my homework
2)i know the answer already, but it was graphically. i want it algaebrically.
3) and if you have nothing else to do but write stupid answers, please write them in your head.

i asked you to solve it not comment on it.
by the way the answer is 0.567143291 till 9 decimal places. so there.

Answer 1

well i think the answer could be :
x = Prodcut log[1]
x = 0.567143

Good Luck.

Answer 2

It's not possible to solve this by transforming the equation. To get a solution of such intricate functions, we can use approximation. Let's use Newton's method:

Step 1:
Transform the equation to have zero on one side:
x*e^x=1 (ln on both sides)
ln x + x = 0
We'll interpret the equation as a function:
f(x) = ln x + x
where the zero of the function, f(x)=0, is our solution

Step 2:
Get the first derivate:
f'(x) = 1/x + 1

Step 3:
Take a wild guess where the zero might be.
Lets guess it's at x=1

Step 4:
Get f(x) and f'(x) for our guessed x:
f(1) = ln 1 + 1 = 0 + 1 = 1
f'(1) = 1/1 + 1 = 2

Step 5:
Create a linear function g which is a tangent to f(x) at the point (x,f(x)), in this case (1,1). It has the slope of f'(1)=2
g(x) = 2(x-1) + 1 = 2x - 1

Step 6:
g(x) is a approximation of f(x), the zero of g(x) is therefore a better guess for the zero of f(x):
0 = g(x)
0 = 2x-1
x = 0.5

Step 7:
Go back to Step 4, this time using our new guess. This should be repeated until the guess doesn't change any more (due to the limited precision a calculator has).

* * *

The third guess after 1 and 0.5 is about 0.56438, which is already pretty close to 0.567143, the value the answerer calling herself or himself "Mistakes/Typos/Miisreads" has determined by using a computer algebra system.

Answer 3

For the best answers, search on this site https://shorturl.im/aw4I5

Hi As Captain Matticus said, x = 1 is a solution. To show that it is the only solution, consider the function: f(x) = e^x - ex (an altered form of the equation you gave replacing 0 with f(x)) f'(x) = e^x - e f''(x) = e^x f''(x) > 0 for all x, so there's only one solution to f(x) = 0, and that's x = 1. If you wanted to use the Lambert W function, you could do: e^x = ex 1 = ex/e^x 1/e = x/e^x 1/e = xe^(-x) -1/e = -xe^(-x) -x = W(-1/e) x = -W(-1/e) x = 1 I hope this helps!

Answer 4

X*(1+X+X^2/2!+...)=1 USE TAYLOR EXPENSION
X+X^2-1=0 TAKE FIRST 2 TERMS
X^2+X-1=0

X_1,2=(1/2)*(-1+-SQRT(1+4))
X_1~=0.618
X_2~=-

Answer 5

No exact algebraic is possible. x has the interesting property that both x and e^x are transcendental numbers.

(It is known from number theory that x and e^x cannot both be algebraic. However, 1 divided by an algebraic number is algebraic. Therefore neither of the numbers is algebraic.)

The equation is equivalent to e^x = 1/x. Because e^x is monotonously increasing, and 1/x monotonously decreasing (x is positive!), there can only be one solution.

Answer 6

Sorry, there is no purely algebraic way of solving this. Essentially, you need to introduce a new function, which will be the inverse of xe^x nad evaluate this new function at x=1. This new function is called the Lambert W-function.

Answer 7

x*e^x=1
Taking log both side
log x + x =0 which cannot be solved hence no value of x satisfy this equation for x>0

Answer 8

With Mathematica I get,

In[2]:=
Solve[x Exp[x]==1,x]
Out[2]=
{{x\[Rule]ProductLog[1]}}
In[3]:=
N[%]
Out[3]=
{{x\[Rule]0.567143}}

So, what is ProductLog[1]?

Using Mathematica Help, we have:

In[4]:=
?ProductLog
"ProductLog[z] gives the principal solution for w in z = w e^w. ProductLog[k, \
z] gives the kth solution."

That is a neat problem! :)

Answer 9

X Exp X

Answer 10

You solve it,it's your homework