Can We Solve A Nonstandard Equation? 😮
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Before even viewing, I smell the obnoxious odor of Lambert's W function.
🙃😜
So did I 😒
BRO SAME-
😂😂
Two is clearly one solution. About two seconds to notice that.
After we guess x=2, we have to show that this solution is unique. Consider the function f(x)=x5^x. Note that for non positive x, f(x) is also non positive. and for positive x, f(x) is strictly increasing (no need to differentiate to see that). so there is a unique solution to our equation.
That's what I thought of instantly.
I got x=2 right away, but I could sense Lambert's W function being used just by looking at the original equation.
First impression: x = 2 is an obvious solution.
If we sketch the graphs y = 5^x and y = 50/x it would seem there can only be the one intersection, therefore x = 2 is the only solution. Now I'll watch the video. I am sure there will be some subtlety that has escaped me. This is a wonderful channel.
It's clever to set y = 5^x, so y=50/x. The exponential function is increasing everywhere, and the rational function is decreasing everywhere, and both functions are continuous for x>0, so they have at most one point of intersection
good point!
@@SyberMath Lambert is ruining my life
Why use Lambert’s W Function when you can easily express that 50 = 2 * 25 = 2 * 5^2, meaning x = 2. It’s a similar process relative to using the W(), but much faster 😑
Because we’ve to confirm that 2 is the only solution. Sometimes we may get a complex solution via Lambert’s function. Taking derivative of x•5^x is to confirm 2 is the only real solution. This step is redundant if we’ve used Lambert’s function.
The prime factorization of 50 is 2*5^2. By 1:1 correspondence x = 2. It’s not “guess and check”.
Yes it is. Your logic is only valid if we were looking for integers x.
I did this. The following step is to know if there are more solutions; which is what Syber do at the end of the video. 👌
x5^x = (2)5²
xe^(xln5) = (2)5²
(xln5)e^(xln5) = (2ln5)5²
(xln5)e^(xln5) = (ln5²)5²
(xln5)e^(xln5) = (ln5²)(e^ln5²)
xln5 = ln5² = 2ln5
*x = 2*
Wonderful! And nary a W in sight! Good for you!
@@josepherhardt164 exactly. also Lambert W function would be useful
I don't quite get why you try to do this dance inside the W function. You can start from the start by saying 50 = 2*25, and recognize that 25 = 5^2. So you rewrite the equation as x * 5^x = 2 * 5^2
Since the function on the left side is trvially increasing, there's only a single solution for x > 0, you can immediately see that x = 2 is the solution.
"t"(tea) or coffee. Ha ha I love it! Better than to "b" or not to be!
🫖
Use Desmos and graph y= x*5^x - 50.
5^(x5^x)=(5^x)^(5^x)=5^50=25^25 therfore, 5^x=25 x=2
I think we can also find the solution factoring the right side od the equation and easily proove the funtion is monotonic finding its derivative as always positive, what causes only one possibile solution found by factoring😁
The function is strictly increasing and continuous, so it's invertible, so the obvious real solution is unique in the reals. But if you want complex solutions also, then lambert W is obviously the correct approach
x.5^x =50
Let x=1, 1x5!=40
Let x=2, 2x5^2 = 50
Just a easy sum, nothing complicated
Until more than 5 terms come
You can see it immediately that x=2.
Seems like this is overcomplicating it to me..
x=W(50ln5)/ln5
x×5^x=50 --> 5×5^x=2×5²
Thus x=2
"t"(tea) or coffee. Ha ha I love it! Better than to "b" or not to be!
☕️
so damn easy
x.5^x=2.25
x.5^x=2.5²
x=2
8:30 Go through all these * hoops * . Not "loopholes".
2 * 5^2 = 50 ho-hum
Since x=2, you are on the upper real branch of the Lambert function. There is only one solution and it is real.
Un un um!!
x = 2
🤑
😜
f=x*5^x is monotonically increasing, so here is only one solution, and it is x=2 obviously, why does video last 10 minutes?
if he replace the 5 with a 4, people should complain that it is not a whole number. The video is titled Can We Solve a Nonstandard Equation and he used a method that should work for any numbers. The video is not titled how to solve this specific equation; it is just how to solve this type of equation.