Solving the Exponential Equation e^x=ex

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[moeberry]

Another way to solve it is using the lambert W function. First divide both sides by e^x and divide both sides by e as well. Then 1/e=xe^-x
Then multiply by -1 on both sides to apply the Lambert W function. Then we get -1/e= -xe^-x
Now we are in the form of applying the Lambert W function after doing so we get W( -1/e)=-x and w(-1/e) is equal to -1 therefore x=1

[ramaprasadghosh]

x=1 is a trivial solution to exp(x-1)= x
Any negative value in RHS ( i.e x< 0) would contradict with positive LHS.
At x = 1 +h
LHS
exp( x-1) = exp(h) = 1 + h + h^2/2 + ..
> 1+h = RHS for all h > 0
Hereby only feasible solution is there at x = 1

[skakistisA]

For the second solution, you also need to argue that the e^x function is either convex or concave.

[andreabaldacci]

I solved it this way. FIrst, since e^x is always positive, if a solution exists, it must be ex>0, meaning x>0. Substituting ex with t, I get ex=t (t>0) and x=t/e. The original equation becomes e^(t/e)=t. Raising both side to the power of e gives e^t=t^e. For t>0, as established above, both sides represents increasing 1:1 functions. Both functions are convex, which means that they can cross at most in one point. This point is t=e, which turns the equation into e^e = e^e. Substituting t=ex=e provides x=1 as the only solution.

[michaelz]

You give the impression of intentionally drawing out the video unnecessarily. The situation is that e^x is convex and y = ex is the tangent line at (1, e). This video basically described that situation in as long of a way as possible, obscuring the picture until the end.

[neilruston]

Simply differentiate again to determine f ' ' and plug in value x=1. If negative, the turning point is a maximum, else a minimum.

[ShubhayanKabir]

Why wouldn't you simply take the double derivative to check whether it's a max or min? That's so straightforward.

[andregoncalves]

I rearranged the equation to find x - ln(x) = 1
(By putting lns in both sides and use the log rules to simplify it)

Then we can see that the function that is in the first member of the equation is always decreasing, so it just have value of 1 once. We conclude that there is only one solution. That solution is 1

[udic]

1:44 why does it matter what kind of critical point it is?
The fact that it is the only one means it is the only solution for f(x)

[JohnRandomness]

Let's check if x = 1 is unique. The derivative of ex is e, and the derivative of e^x is e^x, which is also e when x = 1. The two functions are tangent to each other at x = 1. When x < 1, e^x > x because it approaches with lower slope. When x > 1, e^x > x because it runs off with greater slope.

Consequently, x = 1 is unique.

[Ssilki_V_Profile]

After it was discovered that the only critical point of the function coincides with the already found root of the function, and is a stationary point, the need to define the properties of this critical point disappeared. Regardless of whether this point is an extremum or it is similar to the critical point x^3 - the root found is guaranteed to be the only one.

[oenrn]

Easy way to prove 1 is the only answer:

e^x = ex
Differentiate both sides
e^x = e
From here we get
ex = e
Divide both sides by e
x = 1

[sngmn]

About the 2nd way:
Tangent line eqn: y = ax + b, where "a" is the slope.
The derivative e ^ x is equal to e ^ x, and for x = 1 it is equal to "e", so the tangent line eqn becomes y = ex + b. You still need to prove that b = 0 then everything will fit together.

[carloshuertas]

Another great explanation, SyberMath! I looked at the math problem and I solved for x fast. Thanks a lot!

[iwd]

Needn't you check for f'' not being 0 when x=1? As far as I know from back in the days, you need that in order to conclude a minimum or a maximum for sure.

[scottleung]

All I did was plug in values for x, and 1 was the only solution that worked.

[somwangphulsombat]

Take log this equation it will be x= 1+ ln x and draw the graph of y1=x and y2 = 1+ ln x and find the bisect points.

[shubhamkochar]

Take log on both sides and draw graph so easy

[Ssilki_V_Profile]

If f(x) equal to zero in point, that for - derivative also equal to zero(and that is the only critical point - this root is the unique.

[lazaremoanang]

I don't have the time to watch the video for now.