Solving A Nice Exponential Equation

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #ExponentialEquations #Exponentials
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :

[olerask]

Note to the minimum of x 256^x: The minimum occurs in x ~ -0.18, with minimum value close to -0.07. These are very normal negative numbers. If the equation had negative RHS (f.ex. -0.05), we would find two solutions.
In our case a better argument is (as indicated in another comment):
For positive x the function x 256^x is increasing, and for negative x it only have negative values. Thus x=1/4 is the only solution.

[GabriTell]

When I first saw it I instantly thought "that's gonna be 1/4" ✅️

[dwm]

Perhaps easier to sketch the graphs y = 256^x and y = 1/x. Obviously only one intersection, so the obvious solution is the only solution. Easy, but nice.

[karl]

How about using a similar trick as in the video on "x^(x^3)=36":
instead of raising both sides to some power, as in that older video, take 256^equation, i.e
256^(x*256^x)=256^1
exponent rules give (256^x)^(256^x)=256 BUT 256=4^4, so 256^x=4, done

[ralfbodemann]

I wrote 256^x as e^(8*ln2*x). Then I multiplied the whole equation by 8*ln2 yielding (8*ln2*x)*e^(8*ln2*x) =8*ln2.
Now I can apply the Lambert-W-function which gives 8*ln2*x = W(8*ln2) which can easily be solved for x = W(8*ln2)/(8*ln2).

I dedicate this solution to blackpenredpen.

BTW: This gives the nice side result 1/4= W(8*ln2)/(8*ln2).

[Psykolord]

Given the use of a number that is a perfect square, i started looking at square roots. Square root of 256 is 16. Since we're using square roots, that vives x =0.5. Thus we get 0.5*16, which is 8. Since this failed, lets look at the square root of 16, aka the quartic root of 256.

This gives us x=0.25. The quartic root of 256 is 4, giving us 0.25*4=1.

Therefore, x = 0.25.

[TheNameOfJesus]

Without even opening this page, I used simple inspection to solve it. The numbers I tried were: 0, 1, 1/256, 1/8, 1/4. Five tries, about 2 minutes.

[realcirno]

all solutions are described by W(ln(256)) / ln(256), but on the principal branch the property W(x lnx) = lnx holds, so we have W(ln(256)) = W(4 ln4) = ln(4) so this leads to 1/4 just like in the video

[ruslingcomet]

It is also possible to use the lambert W function, thus giving all possible solutions to this equation. It is fairly easy to apply here, so it just clicked to me.

[goldfing]

x * 256^x = 1
x cannot be zero, since 0 * 256^0 = 0 * 1 = 0 <> 1.
Thus we can divide by x:
256^x = 1/x
Regarding the graphs, we have a monotonically increasing exponential function (base = 256 > 1) and a monotonically decreasing hyperbola for x > 0. Anyway x cannot be negative since 246^x is always positive and 1/x negative for negative x.
There onviously is only one real solution and it is positive.
Try x = 1/4:
1/4 * 256^(1/4)
= 1/4 * sqrt(sqrt(256))
= 1/4 * sqrt(16)
= 1/4 * 4
= 1
So x = 1/4 is the only real solution.

[RexxSchneider]

We can see that f(0) = 0 and f(1) = 256 must bound the interval where f(x) = 1.
Since f(x) = x * 256^x is monotonically increasing for positive x, and is negative for negative x, we can see that the interval [0, 1] must contain the sole solution.
If we can massage the equation into the form a^b = a^c, we can get the solution by equating b and c. But the RHS is a power of 2, so we should first look for a power of 2 as a solution.
Because 0 < x < 1, it motivates us to choose x = 1/2^n so that n is a natural number (we hope).
Now we have 1/2^n * (2^8)^(1/2^n) = 1. Multiplying both sides by 2^n, we get 2^(8/2^n) = 2^n. So equating indices, 8/2^n = n.
But 8/2^n is the same as 2^(3-n)

[j.r.]

Nice equation, except that "guess and check" was a little too easy here! The more formal approach seemed like a bit of a slog.

[JagadeshRaoThalur]

which software or website do you use for animations and graphs? please let me know

[BRUBRUETNONO]

Apart from the guessed "trivial" value 1/4, I made the exercise the hard way which is nice with an excel extension of the Lambert W function.

At the end I use an excel extension I found on the net.
It allows to calculate the Lambert function value W0(x).
W0 is the first branch value of the Lambert function.

E: x.256^x=1
Remark 256=2^8=e^(8.ln2) Then
E <=> x.(2^8)^x=1
<=> x.(e^(8.ln2))^x=1
<=> x.(e^(x.8.ln2))=1
<=> x.8.ln2.e^(x.8.ln2)=8.ln2
<=>
As Lambert W function is [W(y)=x <=> x.e^x=y] then
<=> x.8.ln2=W(8.ln2)
<=> x=W(8.ln2)/(8.ln2)

Approx value of 8.ln2 is 5, 545 and W0(8.ln2) is 1, 386
So approxvalue of W0(8.ln2)/(8.ln2) is 1, 386/5, 545 # 0, 25 # 1/4

Solution= {W(8.ln2)/(8.ln2)}= {1/4}



Excel extension source
exact value approx value
a= 8.ln2 # 5, 545177444
W0(a)= W0(8.ln2) # 1, 386294361
b=W0(a)/a=W0(8.ln2)/(8.ln2) # 0, 25
x W0(x) W0(x)/x
0, 5 0, 351733711 / 0, 703467422
1 0, 56714329 / 0, 56714329
2 0, 852605502 / 0, 426302751
3 1, 049908895 / 0, 349969632
4 1, 202167873 / 0, 300541968
5 1, 326724665 / 0, 265344933
5, 4 1, 370918211 / 0, 253873743
5, 545... 1, 386294361 / 0, 250000
5, 5 1, 381545379 / 0, 251190069
6 1, 432404776 / 0, 238734129
7 1, 524345205 / 0, 217763601
8 1, 605811996 / 0, 2007265
9 1, 67901642 / 0, 18655738
10 1, 745528003 / 0, 1745528

[lucaskenji]

Admiting x=2^p, we have 2^p+8x=0. if x=1/4 we have p=-2 .... 1/4=2^-2 (Only Solution)

[supergaga]

I see a lot of people calling on the lambert W function - i don’t see the need for it, Here’s how I did it, seemed simpler…but maybe I missed something

Suppose x is strictly negative, x256^x is negative, hence x must be positive.

Suppose x = 0, then x256^x is 0, therefore x must be strictly positive.

Set f(x)= ln x +x ln(256) (natural logarithm of the left expression). Since ln is a bijection in R+, solving the equation is simply finding the zeros of f.

This is continuous on ]0;1] and différentiable, strictly growing since 1/x +ln(256) >0 when x>0

f(0.2) is negative, f(0.3) is positive, by Bolzano ‘s theorem there is a unique 0 between .2 and .3

Turns out 1/4 is a natural solution, and it is then the only solution.

Please tell me if there’s a mistake.

[allanmarder]

To avoid guess and check, I used the following Lambert solution: x= -W(8ln(2))/8ln(2) = (approx) -1.386294/-5.54517744 = approx .25 (difference is -

[david_allen]

Simple inspection suggests 1/4 is one solution. Observation that the LHS is monotonically increasing, and RHS is positive suggests that if any other solution exists, it must be between 0 and 1/4. There is none.

[sebastianviacava]

I love to see your videos i m learning a lot

[LePhenixGD]

Has anybody here used the W(x) Lambert function?

I used it like this:



x × (256)ˣ = 1
We know that 256 = 2⁸
x × 2⁸ˣ = 1
//
To solve this problem we need to change the expression such that we can use this property:
aeᵃ= y
a = W(y)

For example:
Ln(x)×x = y
Ln(x)eᴸⁿ⁽ˣ⁾ = y
Ln(x) = W(y)
x = eᵂ⁽ʸ⁾
//

So:
xe⁸ˣᴸⁿ⁽²⁾ = 1
8xLn(2) e⁸ˣᴸⁿ⁽²⁾ = 8Ln(2)
8xLn(2) = W(8Ln(2))
x = W(8Ln(2))/8Ln(2)
x = 1/4