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
@@schukark: Using Lambert W function yields xln4e^(xln4)=ln4. As ln4>0 there is only one solution, i.e. x=W(ln4)/ln4. Using WolframAlpha program to find the x value gives x=½ the same answer that Lev Skomorovsky obtained in a simpler method.
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.
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.
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
You cannot use that method to determine if an equation has only 1 solution. That only works for continuous functions. 1/x is discontinuous at x=0, you would be wrong if the function 4^x was shifted downward then you would have two solutions. Since 1/x is decreasing in two different regions. You got bailed out since 4^x lies above the x axis and therefore doesn’t get into quadrant 3 where 1/x also lies besides quadrant 1.
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
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. 🙂
If the Lambert W Function is an inverse function F^-1, then does the W( ln4) have to be entered into the calculator aa an inverse: ln1/4? It mystifies me what mathematical procedure is being used with W to come up with the answer. In this case the answer is obvious and obviously 1/2.
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.
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.
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.
Thank you, very nice! How do we solve: X squared equals to 2 raised to the X? When I graph these two equations I see three real solutions, but I don't know how to solve it. If you can help it is much appreciated!
The equation x²=2ˣ has two obvious positive solutions, 2 and 4, and one not so obvious negative solution. We can solve the equation using Lambert's W function, using a technique similar to that in the video. x²=2ˣ ⇔x=±√(2ˣ)=±2^(x/2)=±(√2)ˣ ⇔x(√2)⁻ˣ=±1 ⇔x(e^ln √2)⁻ˣ=±1 ⇔xe^(-xln √2)=±1 ⇔(-xln √2)e^(-xln√2)=ln√2 or -ln√2. ⇔-xln √2=W₀(ln √2) or W₀(-ln √2) or W₋₁(-ln 2) ⇔x=-W₀(ln √2)/ln √2 or -W₀(-ln √2)/ln √2 or -W₋₁(-ln √2)/ln √2 The first solution is negative and the other two are positive. In fact ln ½ e^(ln ½)=½ln ½=-½ln 2=-ln √2, and ln ¼ e^(ln ¼)=¼ln ¼=-¼ln 4=-½ln 2=-ln √2, and ln ¼
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?
I truly detest Lambert's W formula. For one thing, as a scientist and engineer I never had to use it, and the other reason is that it looks like a mathematical form of cheating. How about x x^x instead?
if W(ln 4)/(ln 4) becomes 1/2, that would suggest there is some sort of simplification with the rules of W(x) and how it relates to ln. W(ln 4) must equal ln(4)/2 for the answer to make sense.
Hi dear Professor, I have a topic that might interest you and I would like to see a video about it: PROVE THAT FOR EVERY REAL NUMBER NOT having A MULTIPLE OF 2 CONTAINS IN ITS DECOMPOSITION AT LEAST ONE ODD NUMBER
This is trivial (I didn't even watch the video). By visualizing the graphs of 4^x and 1/x, it's clear that x lies between 0 and 1. The graph of 4^x is increasing and 1/x is decreasing, so the intersection point is unique. By inspection, x = 1/2. We check this: 4^(1/2) = 2 and 1/(1/2) = 2. Now I'll watch the video and see all the goodies I missed about the Lambert W function.
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
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
Amazing technique man :)
Cool observation, you didn't prove it's the only root though, f(2)=f(1/x) doesn't give 2=1/x in general (here it did ¯\_(ツ)_/¯)
Awesome observation
@@schukark: Using Lambert W function yields xln4e^(xln4)=ln4. As ln4>0 there is only one solution, i.e. x=W(ln4)/ln4. Using WolframAlpha program to find the x value gives x=½ the same answer that Lev Skomorovsky obtained in a simpler method.
@@schukark But we know there is only one solution by the initial argument in the video.
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.
same, it took me along time to understand it, but once i did, i realized it was very easy very quickly
yes
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.
I’ll have to remember that Lambert function. You use it a lot and it’s helpful.
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
You cannot use that method to determine if an equation has only 1 solution. That only works for continuous functions. 1/x is discontinuous at x=0, you would be wrong if the function 4^x was shifted downward then you would have two solutions. Since 1/x is decreasing in two different regions. You got bailed out since 4^x lies above the x axis and therefore doesn’t get into quadrant 3 where 1/x also lies besides quadrant 1.
@@moeberry8226 I guess you are right!I got carried away by my enthusiasm about finding the solution
@@popitripodi573 no problem but good comment anyways keep it up.
@@moeberry8226 thank you
Nice 😍
I wish you explained how to find value of "W(x)".
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
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. 🙂
Glad it was helpful!
You never really showed how the W function gave the answer.
haha, like it. understand lambert W better and can try to apply it when I meet similar questions in the future.....wonderful
Glad you enjoyed it!
If the Lambert W Function is an inverse function F^-1, then does the W( ln4) have to be entered into the calculator aa an inverse: ln1/4? It mystifies me what mathematical procedure is being used with W to come up with the answer. In this case the answer is obvious and obviously 1/2.
I also used Lambert W function first and then realized this is actually 1/2. :-) hehe
🤪
As W is a multi-valued function, there are complex solutions,
For example:
x= -0.86645...+3.20904...i
1:59
Finally it comes
Nice and brilliant
What is the board application that you are writing on ,.,....?
Please Reply sir 🐱
Sure. It’s Notability
@@SyberMath thanks a lot 😌🤞
The solution of the equation a^x = b/x can easily be found to be x = 1/ln a * W[b* ln a] .
Fantastic. I also didn't know about Lambert's W 😄
Based on inspection alone x = 1/2 works. I suspect Lambert's may provide another solution.
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.
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.
I guessed and checked (didn't have time to work it out)!
Hehe nice!
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.
Thank you, very nice! How do we solve: X squared equals to 2 raised to the X? When I graph these two equations I see three real solutions, but I don't know how to solve it. If you can help it is much appreciated!
The equation x²=2ˣ has two obvious positive solutions, 2 and 4, and one not so obvious negative solution.
We can solve the equation using Lambert's W function, using a technique similar to that in the video.
x²=2ˣ
⇔x=±√(2ˣ)=±2^(x/2)=±(√2)ˣ
⇔x(√2)⁻ˣ=±1
⇔x(e^ln √2)⁻ˣ=±1
⇔xe^(-xln √2)=±1
⇔(-xln √2)e^(-xln√2)=ln√2 or -ln√2.
⇔-xln √2=W₀(ln √2) or W₀(-ln √2) or W₋₁(-ln 2)
⇔x=-W₀(ln √2)/ln √2 or -W₀(-ln √2)/ln √2 or -W₋₁(-ln √2)/ln √2
The first solution is negative and the other two are positive.
In fact ln ½ e^(ln ½)=½ln ½=-½ln 2=-ln √2,
and ln ¼ e^(ln ¼)=¼ln ¼=-¼ln 4=-½ln 2=-ln √2,
and ln ¼
W(ln4) = W(2 * ln2)= W (e^(ln2) * ln2) = ln2
Lambert's function wasn't really needed after all
What about 7^x +8^x= 325?
2
@@SyberMath x≈2.5223
I was a little surprised that you didn't go with your usual x^x. Don't you love that function anymore?! ;-)
I do but I wanted to use Lambert this time! x^x I still ❤️ you
🎉🎉🎉
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?
A respectful suggestion: Less wordy
Thank you!
It is very easy my friend
4^x =1/x
Then 4= (1/x)^1/x
Then 2^2 = (1/x)^1/x
Then 2=1/x
Then x =1/2
I truly detest Lambert's W formula. For one thing, as a scientist and engineer I never had to use it, and the other reason is that it looks like a mathematical form of cheating. How about x x^x instead?
4ⁿ=(1/n)
4ⁿ=n^-1
ln(4ⁿ)=ln(n^-1)
n[ln(4)]=-1[ln(n)]
-1[ln(4)]=(1/n)[ln(n)]
-1[ln(4)]=[ln(n)][e^[ln(n^-1)]
-1[ln(4)]=[ln(n)][e^[-ln(n)]]
ln(4)=[-ln(n)][e^[-ln(n)]]
-ln(n)=W[ln(4)]
ln(n)=-W[ln(4)]
n=e^-W[ln(4)]
n=e^-W[ln(2²)]
n=e^-W[2[ln(2)]] ❤
if W(ln 4)/(ln 4) becomes 1/2, that would suggest there is some sort of simplification with the rules of W(x) and how it relates to ln.
W(ln 4) must equal ln(4)/2 for the answer to make sense.
@@TheEternalVortex42 Aha. Knew there had to be a way since it was such an exact answer.
x = 1/2
0.5
Ln(4) not equal to one
I can't see how W(ln4) =1 ?
W(ln4) does not equal 1. It equals ln2 because
W(ln4)=W(2ln2)=W(ln2*e^(ln2))=ln2
Awesome technique! Still, this was easy enough to solve mentally in less than a second.
.5 ans
And a 7th grader screwed his brain over doing this. Wow
Ln4 is not equal to one.
Hi dear Professor, I have a topic that might interest you and I would like to see a video about it:
PROVE THAT FOR EVERY REAL NUMBER NOT having A MULTIPLE OF 2 CONTAINS IN ITS DECOMPOSITION AT LEAST ONE ODD NUMBER
،👍
This is trivial (I didn't even watch the video). By visualizing the graphs of 4^x and 1/x, it's clear that x lies between 0 and 1. The graph of 4^x is increasing and 1/x is decreasing, so the intersection point is unique. By inspection, x = 1/2. We check this: 4^(1/2) = 2 and 1/(1/2) = 2.
Now I'll watch the video and see all the goodies I missed about the Lambert W function.
x=1/2