A Nice Exponential Equation (e^x=x^e)

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[gmutubeacct]

You could take second derivative to show that f''(x) > 0, and hence x=e is a global (and the only) minima. Hence x=e is the on ki y solution.

[XJWill]

The easiest way is by inspection. Can instantly see x=e is a solution, and the graphs of e^x and x^e are well known, monotonically increasing, and only cross at one point. QED
But if that still is not enough to convince you, just manipulate the equation to get all the x's on one side
e^x = x^e
x*ln(e) = e*ln(x)
ln(e) / e= ln(x) / x
From which it is obvious that the only real solution is x=e

[ekxo]

and there are a lot more complex solutions of the kind e^(-W_k(-1/e)) where W_k is the k-th branch of the Lambert W function, that is the inverse function of this function: f(z) = z e^z. (So, for example: since 1*e ^ 1 = e; then 1 is W_0 (e) (there are many complex branches, 0 returns the real answer, when there is one))

(for those wondering:
e^x = x^e
x = elnx
x/lnx = e
lnx/x = 1/e
-lnx * x^(-1) = -1/e
(-lnx)*e^(-lnx) = -1/e
-lnx = W(-1/e)
x = e^(-W(-1/e))
)

[rcnayak_]

We can have another short-cut and simple solution. Raise both sides to the power (1/xe) so that we get e^(1/e) = x^(1/x). And the result follows by comparison i.e., x = e.

[NEVEU]

I am procrastinating doing my math homework by watching this. You explained this in a proper amount of detail, enjoyed it.

[Megathescientist]

Another way to solve this problem is the following:
let's x=e^(t+1) so that e^(e^(t+1))=e^(e*(t+1)).
This implies that e^t=t+1, which is true only for t=0, as we can see if we write the maclaurin expansion of e^t.
We can conclude then, that the only solution to the problem is x=e^(0+1)=e.

[ferraneb]

After checking x > 0, just substitute x = e^(lnx) on the right side, and then we get e^x = e^(x*lnx) -> x = x*lnx since e^x is increasing and therefore one to one => lnx = 1 => x = e

[pbondin]

My approach was to differentiate both sides with respect of x, so the equation e^x = x^e becomes e^x = ex^(e-1). This means that x^e = ex^(e-1). [since the LH side remains unchanged when differentiated, the two RH sides are equal]. Divide both sides of this equation by x^(e-1), to get x = e. [Incidentally, by inspection, x <> 0, otherwise the original equation becomes 1 = 0. Further x cannot be negative, otherwise the original equation will always have a positive LH and a negative RH]

[benavidezf]

At the fork, x = eln(x), you have e^-1 = ln(x)/x = x^-1 * ln(x) = ln(x) * exp(ln(x^-1) or rewriting:

e^-1 = ln(x) * exp(ln(x^-1) multiply both sides by -1 and we have
-e^-1 = ln(x^-1)*exp(ln(x^-1) now take the Lambert w function of both sides and you have
-1= ln(x^-1) or exp(-1) = x^-1 or 1/e = 1/x and x = e

[somwangphulsombat]

You can plot the graph of y=x and y=e.lnx and find the bisect points.

[catalina]

In the entrance exams of Japanese university, this is one of typical questions. For example, solve 2^x＝x^2. In short, there are many question using the function, lnx/x.

[fdyt]

x = -e * W(-1/e)
Just one real solution, but infinity complex solutions

[Unidentifying]

Tried to do it in my head with purely algebra, great exercise

[JosiahFickinger]

Fun fact. 2 to the 4th power is the same as 4 to the 2nd power. That fits perfectly in this equation.

[uberless]

Awww, he drew the graph by hand. I was looking forward to seeing the Desmos graph in the document over the next week...

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[bollyfan]

Easier way to solve is that domain is x > 0.
We also know that if 0 < x <= 1, then x^e < 1 while e^x > 1
so we get 1 < x
We know that x = e is a trivial solution.
we also know that x^e < e^x when 0 <= x <= 1
so x^e starts out below e^x

if I take the (e * x) th root of both curves, it should preserve the ordering when 1 < x
so we compare x^(1/x) with e^(1/e).
We know that x^(1/x) has a maximum at x = e
Hence x^(1/x) < e^(1/e) when x != e
This means there is only one solution x = e and the curve e^x is tangential to the curve x^e at x = e, and x^e < e^x for all other values of x in the domain 0 <= x

[xenumi]

I feel like by using the fact that x=e gives an absolute minimum on the domain, we did not need to find the limits of infinity on both sides. Everything else > 0 => solution is unique.

[arrrhoo]

Differentiate both sides with respect to x. The left hand side exponential remains invariant after differentiation with respect to x. Equate both right hand sides and solve for x.

[vidiot]

go to desmos and enter the formula y^x=x^y. You get two lines, one being y=x {x>0}, obviously, but also one that looks similar to y=1/x (it is definitely not this line, . Those two lines intersect at exactly (e, e). Every equation that has the form in the video will have two answers, one for each line, except when y or x equals e. You always have the simple answer of y=x, and the complicated answer (except in the case where y = 2 or 4, the only two points with integer solutions) that takes some work to figure out

[farfa]

I followed the original equation until x/lnx = e; so x = e^a and lnx = e^(a-1). So, x = e^(e^(a-1)) = e^a; e^(a-1) = a; e^a/e = a; ae = e^a. Because of how linear and exponential functions grow, a = 1 is the only solution, therefore x = e^1 = e.