2^x = 4x

This problem has two solutions. The second solution, if done by algebra will require a special function called the Lambert W function. In my solution, I used a power series approximation to estimate the W(t)

This is the video I mentioned

[sethdurais]

Something that is seriously overlooked in your videos are your straight lines when dividing the board! I know its just a side note but you have to admit that Mr Prime draws some of the best straight lines! It is extremely satisfying to see 💯

[johnka]

'if the value inside, the argument here, is close to 0' when did you become a physicist? 😆

[davidsousaRJ]

2^x = 4x = 2²x, therefore x = 2^(x-2). Raising the both sides to the (1/(x-2))-th power, we have x^(1/(x-2)) = 2. Note that if we square both sides, we have x^(2/(x-2)) = 4, or x^(2/(x-2)) = 4^(2/(4-2)). By comparison, x = 4 is a solution. The other one is only possible to get using the Lambert function.

[apexgoblin]

blackpenredpen derived a formula for a^x + bx + c = 0 you can use the formula here too!

[domsunny]

That’s is actually so cool, great video man

[dietrichschoen]

Hallo.
Danke für Ihren wiedermal, für mich,
ausgezeichneten Vortrag.
Der Faktor ist: 2, 7726 um x2 zu berechnen.
Mit W(-ln2*1/4)=0, 2148 im w-1-Graphen,
kann man nun den x2-Schnittpunkt finden.
Er ist etwa bei ca. 2, 6 bis 2, 8.
Mit wiederholten Näherungen kann man den
Wert für x2=4 bestimmen.
X2= 2, 7726/ln2= 4.

[BRUBRUETNONO]

Hi,
Thanks for your insterestin problem, that I solved that way here below.
Tell me if you like it.
Of course, I didn't look at your solution.
Greetings and keep up the good job.

BEGIN
Let's name (i) the equation to solve 2^x=4x
Let the function f(x)=2^x-4x from R to R
So the question is to find the roots of f(x)
We can say that f(x) (being the sum of two continuous functions)
is as weel continuous on R.
Let's evaluate the behavior of f(x).
The derivate of f is f'(x)=ln(2).2^x-4
f'(x)=ln(2).2^x-2^2
Then f'(x)=2^2.[ln(2).2^(x-2)-1]
Let's see for what values of x, f is increasing so that f'>0.
So that ln(2).2^(x-2)-1>0
So ln(2).2^(x-2) > 1
So if x verifies
(ii) 2^(x-2) > 1/ln(2)
then f(x) is strictly increasing

Moreover, as
ln(2)>0 (ln(2)#0, 693)
and the function 2^x is strictly positive on R
and the logarythm function is strictly increasing on R+,
we can then take the ln on both sides of inequation (ii)
and it gives
ln[ln(2).2^(x-2)] > ln(1)
ln(ln(2))+ln[2^(x-2)] > 0
(x-2)ln(2) > -ln(ln(2))
x > 2 - ln(ln(2))/ln(2)

Let following equation and value m
(iii) m = 2 - ln(ln(2))/ln(2)
we know as well from inequation (ii)
that 2^(m-2) = 1/ln(2) that we name equation (iv)
We can say that
for x € [m ; +inf[ we have f'(x) > 0 and f(x) is strictly increasing
for x € [-inf ; m[ we have f'(x) < 0 and f(x) is strictly decreasing
Then f(x) has got a minimum value for x=m

Let's evaluate f(x) at -infinite and + infinite.
We can say that for x --> -inf, 2^x --> 0+ and 4x --> -inf
Then for x --> -inf, f(x)=2^x-4x --> +inf

We can say that for x --> +inf, f(x)=2^x-4x is equivalent to 2^x
Then as for x --> +inf, 2^x --> +inf
Then for x --> +inf, f(x) --> +inf

Let's evaluate the minimum value of f, being f(m).
If f(m) is negative we can say that we will have two solutions.

So we have f(m)=2^m-4m we can write as well
from (iii) and (iv) we have

So
So f(m)=2^2.ln[e.ln(2)/4]/ln(2)

As we know ln(2)#0, 693 > 0, then f(m) and ln[e.ln(2)/4] have got the same sign
Then
Let's see if ln[e.ln(2)/4] < 0
Let's see if e.ln(2)/4 < e^0
Let's see if e.ln(2)/4 < 1
Let's see if e < 4/ln(2)
With a calculator we have
4/ln(2)#5, 771
and
e#2, 718
Then e < 4/ln(2) is confirmed and so f(m) < 0

Let's evaluate the value of m = 2 - ln(ln(2))/ln(2)
Let n=ln(ln(2))/ln(2). Then m = 2 - n
We have
1/2 < ln(2)#0, 693 < 1
Then ln(1/2)<ln(ln(2))#ln(0, 693)<ln(1)
Then -ln(2)<ln(ln(2))#ln(0, 693)<ln(1)
Then -0, 693<ln(ln(2))#ln(0, 693)<0
As well 1/1 < 1/ln(2)#1/0, 693 < 1/(1/2)
As well 1 < 1/ln(2)#1/0, 693 < 2

Then -0, 693.1 < n=ln(ln(2))/ln(2) < 0.2
Then -0, 693 < n=ln(ln(2))/ln(2) < 0

Then 2 - 0 < m = 2 - n < 2 - (-0, 693)
Then 2 < m = 2 - n < 2, 693

TO SUM UP,
For f(x) = 2^x - 4x from R to R
f(x) is continuous on R
f(x) --> + inf for x --> -inf and for x --> +inf
f(x) has got a negative minimum value at x = m = 2 - ln(ln(2))/ln(2)
N.B: m has got the following approx value 2 < m < 2, 693
f(x) is strictly decreasing from ]-inf;m[
f(x) is strictly increasing from ]m;+inf[
Then from the "intermediate value therorem" f(x)=0
has got two different solutions x1 and x2
such that x1<m and x2>m

Let's remark that
f(0)=2^0-4.0=1-0=1 then f(0) >0
f(1)=2^1-4.1=2-4=-2 then f(1) <0
f(4)=2^4-4.4=16-16=0 then f(4) =0
Then we can say that 2^x=4.x has two real solutions 0 < x1 < 1 and x2 = 4

As 0 < x1 < 1, we can try to find the value x1 of the kind 1/p as p positive integer
Then f(1/p)=2^(1/p)-4/p
We can see that :
--> 2^(1/p) decreases towards 1+ as p increases.
--> 4/p decreases towards 0+ as p increases.
As 2^(1/p) > 1, then a good approx value for p would be 3 to avoid 4/p going below 1.

Then a good approx value for x1 is 1/3.
A calculator gives us f(1/3)=2^(1/3)-4/3=-0, 073
END

[samzied]

One of the reason I enjoy math is that it transcends our petty egotic drives. Respect for the matter should involve a minimalistic attitude regarding self-promotion when presenting a topic. In any case, one should always make very sure his discourse is blunder-free before thinking he can afford wasting some focus on posing.

[the_real_nayak]

better way - use iterations ; just start with x = 2^x/4 and put x = 0, then keep on putting the result values again in the expession till the value of x is almost equals to the expresison of 2^x/4 ; that'd be your answer

[albajasadur]

Thank you sir. I have two questions. (1) How can we determine the number of real roots ? (2) Can we get the solution x=4 from Lambert W function method ?

[CalculusReviser]

Excellent, clearly explained video :)

[justliberty]

Once you can estimate that the 2nd solution for x is near zero, do a first order Taylor series for 2^x ( = 1 +xln2) and solve that = 4x to directly get x = 1/ (4-ln2) ~ 0.302. Using the 2nd term would get you closer at the expense of solving a quadratic equation.

[yiutungwong]

π in the Riemann Paradox and Sphere Geometry System Incorporated

So Tau = 2π = π^2 = 4

So 2^Tau = 4Tau = 2^4 = 4 × 4 = 16

X can be Solved for 4 and Tau

[DaniDy]

I didnt know there was a formula for the w function wow

[jakehobrath]

Damn YouTube policy!! Now I’ll never know what the flower is called!

[davidbrisbane]

We know that the line y = 4x and the curve y = 2ˣ intersects at one point at most. So, there is at most one real solution.

By observation, x = 4 is a solution and we are done.

[NhaNguyen-cxri]

2^×=4^×
>>2^×-4^×=0
2^×(1-2^×)=0
1=2^×
X=0
X⁰=1

[sriharivithalapur]

If there are multiple solutions, then which solution is achieved by using the Lambert W function? More specifically... In this case can the solution x=4 be achieved using the Lambert W function?

[marianondrejkovic]

If not applying inspection for solution x=4, is it possible to find 4 by algebra via product log function?

[ryansullivan]

5:26 "let's not call it x, let's call it... x"

I had to go back and make sure I heard him right lol