Another Nice Exponential Equation, 2^{x/2}=x

[@fabriziosantin6063]

The other solution comes from the other branch of the lambert W function, in WolframAlpha is lambert w function(-1,-log(2)/2).

[@marklevin3236]

Much easier to work with function ln x /x . Take derivative and we see ghat this function is decreasing for x>e and increasing for 0

[@spelunkerd]

I have a suggestion for another video. This occurred to me while doing the above exercise, and I was able to solve it with the same kind of satisfying difficulty seen in many of your videos. Suppose y=a^x, and the line graph y=x. Depending on the value chosen for 'a', the two curves will meet once, twice or not at all. Solve for "a" such that the two curves are exactly tangent.

[@goldfing5898]

x = 2 and x = 4 are the two obvious integer solutions. Since 2^(x/2) = 2^(1/2*x) = (2^(1/2))^x = (sqrt(2))^x = 1.4142...^x, you can sketch the graph of this exponential function and of y = x and see of there are other intersections (I guess not, since the exponential function is monotonically increasing (convex).

[@goldfing5898]

You could also rewrite 2^(x/2) = 2^(x * 1/2) = (2^x)^(1/2) = sqrt(2^x) = x and then square the equation to 2^x = x^2, and graph y = 2^x and y = x^2 instead.

[@angelmendez-rivera351]

@@goldfing5898 The equation 2^x = x^2 has a different solution multiset than the equation 2^(x/2) = x

[@tontonbeber4555]

2 and 4 are obvious solution. And it's easy to prove that there are no other solution. Just analyze the function f(x)=2^(x/2)-x. All derivatives with order >=2 are positive, so there is only 1 minimum between 2 and 4 and function is monotonic on both sides of that minimum, so only one solution on each side of the minimum.

[@mcwulf25]

Easy to miss that second solution. I did it without ln. Raise both sides to power (1/X). The LHS becomes 2^(1/2), or √2. So x ^(1/x) = ✓2 Looking at the shape of the function shown here we can find a single minimum and then know there are two solutions.

[@mcwulf25]

(maximum)

[@walterwen2975]

Another Nice Exponential Equation: 2^(x/2) = x; x = ? 2^(x/2) = x; 2^x = x^2, x ≠ 0 Trail-and-error method: Case I: x > 0 x = 1.0, 2^x = 2^1 > x^2 = 1^2 = 1; x > 1 x = 3.0, 2^3.0 = 8.0 < 3.0^2 = 9.0; 3 > x > 1, In-between x = 2.0, 2^2.0 = 4.0 = 2.0^2 = 4.0; Confirmed x = 3.5, 2^3.5 = 11.3 < 3.5^2 = 12.3; x > 3.5 x = 4.5, 2^4.5 = 22.6 > 4.5^2 = 20.3; In-between x = 4.0, 2^4.0 = 16.0 = 4.0^2 = 16.0; Confirmed Case II: x < 0 x = - 1.000, 2^(- 1.000) = 0.500 < (- 1.000)^2 = 1.000; 0 > x > - 1.000 x = - 0.750, 2^(- 0.750) = 0.595 > (- 0.755)^2 = 0.563; - 0.750 > x > - 1.000 x = - 0.760, 2^(- 0.760) = 0.591 < (- 0.760)^2 = 0.578; - 0.760 > x > - 1.000 x = - 0.766, 2^(- 0.766) = 0.588 ≈ (- 0.766)^2 = 0.587; Proved Final answer: x = 2, x = 4 or x = - 0.766 The calculation was achieved on a smartphone with a standard calculator app

[@fubaralakbar6800]

Exponent rule y^(a/b)=bth root(y^a) therefore: 2^(x/2)=sqrt(2^x)=x Now square both sides for: 2^x=x^2 Now divide both sides by x to get 2=x. It's one of the solutions anyway.

[@sanshiptaassimplified13]

Bro what after dividing botht the sides by 2 i gues you just cancelled the x in the denominator with the power 😂😂

[@fubaralakbar6800]

@@sanshiptaassimplified13 Umm I never divided by 2? I used exponent rule and then divided by x

[@sanshiptaassimplified13]

@@fubaralakbar6800 sorry i emant x instead of 2 but the point still stands what after that and the rest of my comment😄😄😅

[@fubaralakbar6800]

@@sanshiptaassimplified13 Okay but my point stands too. The math is valid, look it up.

[@sanshiptaassimplified13]

@@fubaralakbar6800 bruh💀

[@yeetboy4431]

I’m such a genius it’s 2 omg 🤯

[@thidasvinnath8017]

very good explaination but there's a way to solve without logs (only one value not all values) 2^((x)/(2))=x 2^((x)/(2)*(1)/(x))=x^((1)/(x)) 2^((1)/(2))=x^((1)/(x)) the known values somewhat corospond to the variables here with that assumption x=2 is in fact a solution

[@someperson188]

f(x) =: ln(2)x - 2 ln(x), for x > 0. f'(x) = ln(2) - 2/x. So, f' has one zero. Thus, f has at most two zeros. Since f(2) = f(4) = 0, the only zeros of f are x = 2, 4.

[@syedmdabid7191]

Valde valde facile! X= 4 responsi.

[@PhoenixR372]

How can one also solve this exponential eqn similarly like above 16^x=x^2

[@SyberMath]

Hi there! See you at the PREMIERE for the solution (14:00 or 2pm in UTC) kzbin.info/www/bejne/b2HYnaOoYsSAfNU 😉😍

[@dariosilva85]

It is actually not correct to say: 2^(x/2) > 0 implies x > 0.

[@moeberry8226]

It is actually correct to say it. 2^(x/2) is always positive so if the line y=x is equal to that exponential function it must also be positive.

[@dariosilva85]

@@moeberry8226 But you are saying a lot more with your words, than this line: 2^(x/2) > 0 implies x > 0.

[@moeberry8226]

@@dariosilva85 that’s not what Syber Said. It is not the case that 2^(x/2) is positive implies x is positive in general, the concept is because we have it SET EQUAL to x itself that it must be greater than 0. Consider two functions f and g in this case f is equal to 2^(x/2) which always lies above the x- axis. The second function g in this case is x, if they are equal that means they must have the same coordinates at any point of intersection. That means their y values must be the same at those points but since 2^(x/2) can only take positive y values then that implies the second function g which is x can only take positive y values as well but y=x for g therefore x>0.

[@dariosilva85]

@@moeberry8226 The correct statement is: 2^(x/2) = x implies x > 0

[@angelmendez-rivera351]

@@dariosilva85 The correct statement is "for all real x, [2^(x/2) = x & 2^(x/2) > 0] implies x > 0."

[@Wangkheimayum421]

x=2 and x=4 will work perfectly

[@rakenzarnsworld2]

x = 2 or 4

[@broytingaravsol]

x=2, 4

[@mathswan1607]

x=2 or x=4

[@giuseppemalaguti435]

x=e^(-W(ln1/sqrt2))

[@goldfing5898]

Using the Lambert W function does not mean that simplifying is not necessary 🙂

[@giuseppemalaguti435]

​@@goldfing5898 ... Is the 2 method of syber

[@angelmendez-rivera351]

@@giuseppemalaguti435 You need to simplify the expression you provided.

[@antenym8947]

@@angelmendez-rivera351 you only need to simplify though if it isnt the maximum simplification in exact form (which in this case it isnt xD).

[@angelmendez-rivera351]

@@antenym8947 Exactly.

[@3O_9-j4v]

X=2; 4

[@barakathaider6333]

👍

[@plamenpenchev262]

x =2 & 4

[@forcelifeforce]

x = 2 OR 4

[@plamenpenchev262]

@Bruce Norris I put a thumb up to you, but we in Bulgarian just say "this and that". I think the problem was around but in other form.

[@angelmendez-rivera351]

@@plamenpenchev262 In mathematics, the correct connective to use is OR. Your mother language is irrelevant. It cannot be the case that x = 2 AND x = 4, because by the transitive property of =, it follows that 2 = 4, which is false.

[@plamenpenchev262]

@Angel Mendez-Rivera it is very funny to explain logical function to a 61 year chemistry professor that teaches about them his students in Chemometrics class. You just reminded me about my teachers in elementary school.

[@PROTAEQUESO98]

x = 2,4