Solving An Exponential with Lambert's W

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[salvatorecharney]

W(ln4)/ln4 can be rewritten very easily as 1/2. W(ln4) itself can be rewritten as W(ln[2^2]), then W(2ln2), then W([ln2]e^[ln2]), which just becomes ln2 because ln2 is the x in that function. Then when replacing W(ln4) with just ln2, you have ln2/ln4, or ln2/(2ln2), which is just 1/2. Idk if anyone else knew that but I’ll just say that anyways haha

[levskomorovsky]

Such tasks are solved on the count of one or two!
4^x =1/x
(4^x)^1/x = (1/x)^1/x
2^2 = (1/x)^1/x
2= 1/x
x=1/2

[justabunga]

That method works if you use Lambert W function. There is another method since 4 is a power of 2. We can raise both sides by 1/x, which is (1/x)^(1/x)=2^2. Comparing the same base and the same exponent 1/x=2, so x=1/2.

[spelunkerd]

It took me way longer than it should have to understand Lambert W, I kept stumbling on what x is. What finally saved me was to go back to the analogous situation of y=e^(x) and its relation to the inverse function of y=ln(x). The key for me was to remind myself that x in those two equations is not the same, it is simply the input to each function. So, to do the inverse W function, effectively you're saying, you give me y in y=xe^x, and the W function calculator will spit out x from that equation. Looking back, it's so obvious it is hardly worth mentioning, which may be why it is hard to teach.

[vighnesh]

I wish you explained how to find value of "W(x)".

[popitripodi]

I saw that 1/2 was a solution and by the fact that the one function is increasing and the other is decreasing that was the only solution!!! I am not familiar with the lambert’s W function and I 😍 ❤❤❤it

[gdtargetvn]

1/x is strictly decreasing and 4^x is strictly increasing

y = 4^x lies entirely above y = 0 as 4^x > 0 so it has 1 intersect with the other branch of 1/x which is in the positive side

By inspection, x = 1/2

[Paul-]

I’ll have to remember that Lambert function. You use it a lot and it’s helpful.

[JXSJ]

You never really showed how the W function gave the answer.

[GillesF]

Your video provided me with VERY useful clarifications on the W Lambert function: the key to control such a function is the logarithms and the properties on logarithms, mainly. Logarithms are a powerful mathematics tool! Thank you. 🙂

[tylergordon]

As W is a multi-valued function, there are complex solutions,

For example:
x= -0.86645...+3.20904...i

[levskomorovsky]

Another option:
x ln 4 = ln(1/x)
ln 4 = x^(-1) ln ( x^(-1))
ln 4 = e^ln x^(-1) ln ( x^(-1))
W (ln 4) = W [e^ln x^(-1) ln (x^(-1))
W (ln 4) = - ln x
x = e ^(-W(ln4))
And this is according to the formula: e^nW(x) = [x/W(x)]^n, if n = -1.

[snejpu]

I also used Lambert W function first and then realized this is actually 1/2. :-) hehe

[poloniante]

We can solve this without using the LambertW function:
4^x = 1/x, where "x" is not 0.
(4^x)^(1/x) = (1/x)^(1/x)
4^(x*1/x) = (1/x)^(1/x)
4 = (1/x)^(1/x)
let y = 1/x
4 = y^y
2^2 = y^y, that's y = 2
1/x = 2
x = 1/2

check:
4^1/2 = 1/(1/2)
2 = 2.

But I think that method and using LambertW is so useful and helpful.

[henry_dschu]

haha, like it. understand lambert W better and can try to apply it when I meet similar questions in the

[renesperb]

The solution of the equation a^x = b/x can easily be found to be x = 1/ln a * W[b* ln a] .

[renesperb]

Here it is easy to guess the solution x= 1/2, which is the only real solution. If you would choose 3 instead of 4, then you have to the Lambert function.

[jim]

Based on inspection alone x = 1/2 works. I suspect Lambert's may provide another solution.

[Questiala]

My attempt before watching:

Rewrite 4^x as e^xln(4)
e^xln(4)=1/x
multiply by xln(4) we get xln(4)*e^xln(4)=ln(4)
Take lambert on bother sides
Xln(4)=W(ln(4))
X=W(ln(4))/ln(4)
THis right?

[MathOrient]

Fantastic. I also didn't know about Lambert's W 😄