2^x = 4x 

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

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 💯

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.

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

That’s is actually so cool, great video man

Love your work man:)

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.

Excellent, clearly explained video :)

The most ASMR voice ever!!!

Just curious. Instead of using the appropriate approach you did, can’t you just use the actual lambert W function? You’ve shown it a few times. That would get you the number you’re looking for regardless of how close to zero the answer is or not, wouldn’t it?

You celebrate Math. Great to watch.

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

I literally love his videos ❤

love from Dubai!

Is there any way to find those other lambert w function branches without using product log calculators?

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

Why did it blur the phi

Congrats from France

Damn RU-vid policy!! Now I’ll never know what the flower is called!

Pretty sure your voice got muted or something when you were talking about phi sir. Maybe an issue with youtube?

Great video man

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 ?

Really awesome, many thanks, Sir!

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

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?

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

9:19 "Come on!"

Of course, you could just use the solver function on your calculator, but where's the fun in that 🎉

Is there any way to get a value for x in the equation 3^x^x = 10? Just like to know because the only way I gotten a value was from a graphing calculator.

Gostei muito e obrigado

why did you round to 0.309? The actual answer according to wolfram alpha is 0.3099 which rounds to 0.310

Don't have access to the internet, but can watch on youtube 🎉🎉

How quickly would we have gotten to a reasonable answer just using Newton’s method from the start?

just divide both sides by zero

Did it by contoured method and solutions coming are 4 and approximately 0.309905

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

Why does the W function not give x=4 as the solution?

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

X=W(Ln(4th root of 2))/-Ln(2)

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 f(m)=2^m-2^2.m=2^2.[2^(m-2)-m] from (iii) and (iv) we have f(m)=2^2.[1/ln(2)-2+ln(ln(2))/ln(2)]=2^2.[1-2ln(2)+ln(ln(2))]/ln(2) So f(m)=2^2.[ln(e)-ln(2^2)+ln(ln(2))]/ln(2) 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)

Prove the MacLaurin expansion of the lambert function next

"1? 2? 3? 4? Ye 4."

W function

Omg… Is it a BLACKMATH???

four

X=4 is the answer Take log base 2 in both side and solve further

2^4=4*4 x=4 I didn’t graph use a calculator or anything I did it in my head.

❤

4

By inspection, x=4 as 2^x=2⁴=16 and 4x=4(4)=16

x=4

This use of a case specific function to get a numerical approximation seems to support my suspicion that maths is a branch of engineering ;)

There is a math error at 9:10. You forgot to divide the LHS by 4. So the solution is not -W(-ln(2))/ln(2) but -4 * W(-ln(2)) / ln(2)

X =4. I did it in my mind.😅

How’s the approximation of the function found looks like some Taylor series stuff

4