Instead of using the product log function at y*4^y=1 At 1:42 The equation 2^(1+x) = - x If we set x = -k We get that 2^(1-k) = k 2/2^k = k or 2 = k*2^k Which can also be written as 1*2^1 = k*2^k So We can say by symmetry k = 1 or use product log as the bases are same W(1*2^1) = W(k*2^k) So k=1 or x = -1
I love that technique for plotting the two curves to show that they only intersect once in this case, showing only one real solution. But what about complex solutions? I can think of situations where the curves don't intersect at all, yet there are complex solutions.... For example y=x^2 +2 and y=1. Is there some way to quantify how many of those might be present?
I was so sure you'd solve it using your favourite trick, by getting the equation into the form a^a = b^b and inferring that a = b is a solution. It's actually possible to do this. Raise both sides to the power of 2^(-x) and you get 2^2 = (2^(-x))^(2^(-x)).
4^(2^x) = 2^(-x) 4^(2^x) = 1/(2^x) Substitute 2^x = t 4^t = 1/t A Quick graph Shows that there is only one real solution at about t = 1/2. Indeed, 4^(1/2) = sqrt(4) = 2 and 1/(1/2) = 1*(2/1) = 2, so t = 1/2 is a solution. Resubstitute: t = 1/2 2^x = 1/2 2^x = 2^(-1) x = -1 is the only solution for x, because exponential functions like 2^x are monotonical and thus injective.
I couldn’t help but think of that statement you said : that for two functions f(x) and g(x) , that if f(x) is always increasing and g(x) is always decreasing, that they are going to only intersect at a single point . Is the more accurate statement that they would intersect AT MOST a single point? I ask because Let f(x) = e^x + 1 (always increasing) g(x) = -e^x + 1 (always decreasing) Interects at 0 points ?
Yes. At most one solution. But there are no solutions at all if the minimum value of the increasing function is bigger than the maximum value of the decreasing function
To be clear, "increasing" and "decreasing" are not enough for this, you need to add "strictly" in front; otherwise, f(x) = 0 is both increasing and decreasing and has infinite intersections with itself. However, the intersection between an increasing and a decreasing function will always be an interval.
If f(x) = 4^(2^x) and g(x) = 2^-x then ( f - g ) / (x+1) has no roots, is concave up and everywhere positive. And you can observe that graphically. So you know that x=-1 is the one and only root
GOOD AFTERNOON, SIR ： THIS QUESTION IS TOO SIMPLE! JUST TAKING (2to the power of x)th root ON BOTH SIDES, THEN WE COULD GET THE ANSWER RIGHT AWAY, NAMELY x = -1
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Instead of using the product log function at y*4^y=1
At 1:42 The equation 2^(1+x) = - x If we set x = -k
We get that 2^(1-k) = k
2/2^k = k or 2 = k*2^k
Which can also be written as 1*2^1 = k*2^k
So We can say by symmetry k = 1 or use product log as the bases are same W(1*2^1) = W(k*2^k)
So k=1 or x = -1
Just went on RUclips and managed to catch one of these at premiere. Thank you for your content, it's helped me a ton!
Glad to hear that
This kind of problem can be solved by using Lambert W function.
I love that technique for plotting the two curves to show that they only intersect once in this case, showing only one real solution. But what about complex solutions? I can think of situations where the curves don't intersect at all, yet there are complex solutions.... For example y=x^2 +2 and y=1. Is there some way to quantify how many of those might be present?
I was so sure you'd solve it using your favourite trick, by getting the equation into the form a^a = b^b and inferring that a = b is a solution. It's actually possible to do this. Raise both sides to the power of 2^(-x) and you get 2^2 = (2^(-x))^(2^(-x)).
Wow!
@@SyberMath I would never have thought of doing that before watching your channel
Expodential and a little bit of calculus!!!I liked it!! An easy one!!❤❤❤❤❤
Glad you liked it!
Thanks Professor, your explanation always over excellent, thanks again.
4^(2^x) = 2^(-x)
4^(2^x) = 1/(2^x)
Substitute 2^x = t
4^t = 1/t
A Quick graph Shows that there is only one real solution at about t = 1/2.
Indeed, 4^(1/2) = sqrt(4) = 2 and 1/(1/2) = 1*(2/1) = 2, so t = 1/2 is a solution.
Resubstitute:
t = 1/2
2^x = 1/2
2^x = 2^(-1)
x = -1
is the only solution for x, because exponential functions like 2^x are monotonical and thus injective.
Got it!
I'm sure 4:17 it's not 4^(y) = - y but it's 4^(y) = 1/y after substitution. Anyways, great job, thanks for that!
I couldn’t help but think of that statement you said : that for two functions f(x) and g(x) , that if f(x) is always increasing and g(x) is always decreasing, that they are going to only intersect at a single point .
Is the more accurate statement that they would intersect AT MOST a single point? I ask because
Let f(x) = e^x + 1 (always increasing)
g(x) = -e^x + 1 (always decreasing)
Interects at 0 points
?
Yes. At most one solution. But there are no solutions at all if the minimum value of the increasing function is bigger than the maximum value of the decreasing function
You're right!
To be clear, "increasing" and "decreasing" are not enough for this, you need to add "strictly" in front; otherwise, f(x) = 0 is both increasing and decreasing and has infinite intersections with itself. However, the intersection between an increasing and a decreasing function will always be an interval.
If f(x) = 4^(2^x) and g(x) = 2^-x then
( f - g ) / (x+1) has no roots, is concave up and everywhere positive. And you can observe that graphically.
So you know that x=-1 is the one and only root
👍
X=-1.
GOOD AFTERNOON, SIR ： THIS QUESTION IS TOO SIMPLE! JUST TAKING (2to the power of x)th root ON BOTH SIDES, THEN WE COULD GET THE ANSWER RIGHT AWAY, NAMELY x = -1
Hi! Are you sure?
W(ln4) /ln(1/2)=-1
x = -1
x = -1
x = -W(2ln2)/ln2
No, you didn't find a solution, you guessed a solution. If you can't explain the process to a computer (an algorithm) it's not a solution.
Computers cannot understand everything. That's why we are better than computers
x =-1
x=-1