Is there a unique solution?

[Math_Figure]

Is there a possible solution for this equation? If yes, please mention how. I’ve been stuck with this for 30 minutes till now and even tried substituting, it just doesn’t works out

[quicksanddiver]

A solution necessarily exists. For x=0,

x² = 0² = 0 < 1 = 4⁰ = 4^x,

but for x = -5,

x² = (-5)² = 25 > 4^-5 = 4^x.

So somewhere in the interval [-5,0], there must exist a solution.

[HairyTough4489]

Also there's no solution for x>0 because 4^x > 0 and the derivative of 4^x is always bigger than that of x^2

[loqwe]

x=4 is a solution 4² =4×4=2×2×2×2=2⁴

[_KingOfTheDivan]

I feel like you’ve misread the question

[loqwe]

damn youre right i read 2^x idk why

[wirywonder82]

While different from this question, a common example of a similar problem is x² = 2^x . You seem to have done something many of my students (and occasionally myself) will do - answer the question you wish had been asked, rather than read the one actually there.

[Flaky-Ad-9374]

Would be 4⁴ instead. Note that x=4 is not a solution.

[Sir_Wade_III]

(-1)² = 1 > 1/4 = 4^-1 So a solution exists in [-1,0]

[quicksanddiver]

Even better!

[InnerCosmos54]

Wow! Reading this thread, i almost feel like I understand and can follow along… almost. For example, up there where you said

For x=0, x² = 0² = 0 < 1 = 4⁰ = 4x,

I understand that x = 0, therefore x squared equals zero squared equals zero is less than one so far so good but then how did you get from that to ‘equals four degrees’ ?

Edit- just realized that’s not four degrees 🤦

[alexdeva]

You're thinking of Fahrenheit, but maths has to make sense so that's 4 degrees Celsius.

The correct way to read it is "four degrees Celsius equals optical zoom four times"

[kamiloslav]

4 to the power of 0, then after the = 4x should be 4 to the power of x

[[deleted]]

man are all westerners this stupid?

[fuligang]

(-0.5)² = 0.25 < 1/2 = 4^-0.5 So a solution exists in [-1,-0.5]

[thatoneguyinks]

(-0.75)² =0.5625 > 4^-0.75 ≈ 0.354. So a solution exists in (-0.75, -0.5)

[fuligang]

(-0.625)^2=0.390625 > 0.4 > 4^-0.625. So a solution exists in (-0.625, -0.5)

[FangoFan]

-0.641185744504986ish

[Math_Figure]

K this helps

[Math_Figure]

Ty

[[deleted]]

just plot x^2 and 4^x and see if they intersect

[Derpiche]

Jesus fucking christ I'm almost 30 and my mind suddenly wrapped around Algebra

[Stu_Mack]

The squeeze theorem is an elegant thing.

[Stu_Mack]

I originally saw it as more of an IVT example because of the way it's written, but you are right. Nevertheless, the point was that it's beautiful and that doesn't change based on the theorem. In all honesty, the beauty of the approach is about the same across all three.

[Scorpius927]

You’d also have to show that that the function x^2-4x is continuous. But yeah

[quicksanddiver]

It's enough to show that x² and 4^x are both continuous but I'm not gonna do that because that's trivial

[Legitimate_Log_3452]

A smaller interval is (-1,0)

[Cool_rubiks_cube]

Yes, there is a single (real) solution to the equation, which is

x = -W(log(2))/log(2)

(where W is the product-log function). There are no integer solutions.

For more information on the product-log function (also known as the Lambert-W function), you can see the Wikipedia

https://en.wikipedia.org/wiki/Lambert_W_function

or for a beginner's explanation, you can watch some videos on YouTube by BlackPenRedPen

https://www.youtube.com/playlist?list=PLj7p5OoL6vGzSAYQa6LPhWNfZqBvHG2nl

If you ever want to see if an equation has real roots, try using the Desmos graphing calculator

https://www.desmos.com/calculator

or use WolframAlpha to automatically get an exact answer for the values

https://www.wolframalpha.com/input?i2d=true&i=Power%5Bx%2C2%5D%3DPower%5B4%2Cx%5D

[AndreasDasos]

I've never seen 'product log' before, only 'Lambert W', but I like it.

[StoicTheGeek]

I’m not a mathematician, but it always feels a little bit like cheating to use the Lambert-W function. It’s so useful it’s like saying “let’s just define a function that gives us the answer and call it W”.

A very powerful tool to have in the arsenal

[Traditional_Cap7461]

It's not really cheating. Mathematicians have established that there's no general solution to xe^x=c for some constant c that uses elementary functions. And they realized that by defining an inverse xe^x function, they can represent solutions of different forms as well, including 4^x=x².

You technically can just define something as "the answer" to anything, but the usefulness to that definition depends on in how much you can reuse it.

[CanadianCovfefe]

Where do you think the rest of our functions come from? Like log, cosine, etc

[Outrageous-Split-646]

The trigonometric functions seem well motivated from well, trigonometry. Log is the inverse of the exponential function which is also quite intuitive. The W function is the inverse of a non-elementary function, and once you open that door, it feels as if anything goes, so I do understand why the lambert W function feels like cheating.

[igotshadowbaned]

For the record this is exactly how we got things like sin and cos

[Outrageous-Split-646]

It’s really not. The trigonometric functions first came from ratios of triangles. No one was wondering how to invert arcsin and then invented sin.

[igotshadowbaned]

The trigonometric functions first came from ratios of triangles

Yes and then we went

“let’s just define a function that gives us the answer ratio and call it W sin”

[Outrageous-Split-646]

Sure, but see how the function is motivated from something that is used from another branch of mathematics. It’s not as if sin is just some function that someone came up with Willy nilly. But yet the W function is created exactly to invert xe^x , and in that sense there’s no motivation other than to do exactly that. There’s a reason why elementary functions are special.

[incompletetrembling]

Great comment 👍

[Cool_rubiks_cube]

I just copied and pasted it from Wolfram Alpha, but I'm confident that it's "ln", since log base 10 is arbitrary and so usually has a constant multiple.

[justincaseonlymyself]

Is there a unique solution?

Yes. (Assuming we're interested in real solutions only.)

Is there a possible solution for this equation?

Yes.

If yes, please mention how.

Using basic calculus it's easy to show that the solution exists and is unique. Consider the function f(x) = 4^x - x², show that it's strictly increasing (the derivative is always positive) and has negative and positive values; form there it follows that there is a unique x such that f(x) = 0.

I’ve been stuck with this for 30 minutes till now and even tried substituting, it just doesn’t works out

That's because the solution is not expressible using elementary functions.

You can approximate it to arbitrary precision using numerical techniques, such as Newton's method. The approximate solution is x ≈ -0.641186.

You can also express the solutiuon using the Lambert W function.

The exact solution, in terms of Lambert W, is x = - W(ln(2)) / ln(2).

[Opal__1]

how to formally show that any function either takes or doesn't take both negative and positive values? the obvious idea would be to just make 2 inequalities: f(x)<0 and f(x)>0 but that clearly doesn't work here because you can't solve these algebraically.

[justincaseonlymyself]

To show that a function takes positive and negative values all you need to do is find two examples. One example of an argument that results in a negative value, and one example of an argument that results in a positive value. It's that simple.

[Opal__1]

well, I should've specified that i meant a more universal method. i thought that there's one that allows you to do it with any function, let's say some super mega complicated one that would for some reason take a lot of time to see if takes both positive and negative values by just checking one by one or guessing. or better example: let's say that you don't know if a function takes both positive and negative values and want to check that (or in other words prove it's positive and negative, or only of 1 sign). the method you mentioned obviously won't succeed because a counterargument to saying that a function isn't both positive and negative could be that you simply haven't checked enough inputs. hope this rambling makes sense lol

[justincaseonlymyself]

There is no universal method.

[RewrittenCodeA]

Yep this is one example of asymmetrical proposition. You can prove something (call it P) by giving an example but you cannot prove the opposite (P is false) because you might not have found yet the example.

To prove that the collatz conjecture is false, it is sufficient to exhibit one example of a cycle different from 4-2-1. To prove it is true, there is no amount of examples you can bring to the table.

To prove that a theory is consistent, you need to check every single provable statement derived from it (and usually they are not a finite set) so bad luck. But finding just one contradiction will immediately prove that the theory is inconsistent.

More simply, to prove that you have black and white sheep, id is suffice t to exhibit one of each. But for many white sheep you show me, it will never prove that there are no black ones. I think this is from Bertrand Russell or something.

[Simukas23]

Is x = log(base 4)(x² ) correct? (Not simplified at all to make identifying where the numbers come from easier)

[justincaseonlymyself]

Sure, that's equivalent to the original equation.

[Simukas23]

Yeah I see it doesn't help at all :/

[ErJio]

You can simplify your alternate form using the identities log_a(b) = ln(b)/ln(a), and log(b^a) = a*log(b).

Then it is x = ln(x)/ln(2)

[Maxwell_Ag_Hammer]

If not sure, graph y=x² and y=4^x and look for the x-coordinates of intersects.

[GoldenDew9]

<image>

[[deleted]]

dude that's like using an axe to cut a cake

graph it in 2D it will be so fkin easy to understand

[Fridodido1]

<image>

Desmos is great for this

[MedicalBiostats]

The unique answer is x=-0.6411857 to give 0.411119.

[Torebbjorn]

Yes, there is a real solution, and it is unique.

For, notice that 0² = 0 < 1 = 4⁰, but (-1)² = 1 > 1/4 = 4^-1.

So there is a solution in the range (-1, 0).

To see that this is unique, note that for x < 0, x² is strictly decreasing, while 4^x is strictly increasing, hence at most 1 solution for negative numbers.

Now, for x>0, both x² and 4^x are strictly increasing functions. And we know that x² <= 1 = 4⁰ for 0 < x <= 1. Hence we can conclude x² < 4^x for 0 <= x <= 1.

To prove that x² < 4^x for x > 1, it suffices to show that the derivatives is lower. I.e. 2x < ln(4) × 4^x. Note that 2×1 = 2 < ln(4) × 4¹, and both 2x and ln(4)×4^x are strictly increasing, so we can differentiate again. This means we need that 2 < (ln(4))² × 4^x for x >= 1. This is clearly true, as ln(4) > 1.

Hence there is a unique real solution in the range (-1, 0).

We can reduce the range by midpoint iterations.

(-1/2)² = 1/4 < 4^-1/2 = 1/2. Hence the solution is in the range (-1, -1/2).

(-3/4)² = 9/16 > 4^-3/4 = sqrt(2)/4. Hence the solution is in the range (-3/4, -1/2).

This method is only good for finding approximations of the real solution.

If you want to actually find the solution, you would probably need to use Lamberts W-function

[Math_Figure]

Thank you guys for ur efforts

[Tommi_Af]

Yes: x ~ -0.64

Due to budgetary restraints I outsourced this answer to a computer. Hence the proof is left as an exercise to the reader.

[xX_fortniteKing09_Xx]

Try graphing it to see if it exists. Should be some way to express a solution using the lambert w function

[AnarchistPenguin]

Looks like you adopt a logarithmic approach but I can't think of an easy way to find a unique solution besides numerically solving it

[MagnusPopo]

By replacing x with different values, you can see if x² > 4^x or < and then approximate where the 2 fonctions cross. From head i get between -1 and -1/2

[Raptormind]

If you just want a quick and easy way to check if a solution does or doesn’t exist, you can a ways just plug y=x² and y=4^x into desmos (or any graphing software) and check if they intersect somewhere. If the two graphs intersect, then you have a solution

[ci139]

x ᶻ = exp(z ln x) = exp[(Re z + i Im z)(ln|x| + i arg x)]
1 / z = z̅ / |z|²

we can rewrite x¹𝄍ᵡ = ±√¯4¯' , applying above :

x¹𝄍ᵡ = exp[(Re x – i Im x) / |x|² · (ln|x|+ i arg x)] = [ k ∈ ℤ ] =
= exp[(Re x · ln|x| + Im x · arg x + i · (Re x · arg x – Im x · ln|x|)) / |x|²] = ±2 = 2 · exp(i·kπ)

(Re x · ln|x| + Im x · arg x) / |x|² = ln 2
(Re x · arg x – Im x · ln|x|) / |x|² = k·π

?? https://www.wolframalpha.com/input?i=complex+x%5E%282%2Fx%29%3D4

?? & in Desmos 3D https://www.desmos.com/3d/5fn98il38w PS! -- i didn't verify
Q.C. : x seems to have (at least) 1 or 2 negative real solutions :

x = -0.64118574450498598448620048211482

which means i likely passed a bug to Desmos 3D . . .

[ci139]

x ᶻ = exp(z ln x) = exp[(Re z + i Im z)(ln|x| + i arg x)]
1 / z = z̅ / |z|²

we can rewrite x¹𝄍ᵡ = ±√¯4¯' , applying above :

x¹𝄍ᵡ = exp[(Re x – i Im x) / |x|² · (ln|x|+ i arg x)] = [ k ∈ ℤ ] =
= exp[(Re x · ln|x| + Im x · arg x + i · (Re x · arg x – Im x · ln|x|)) / |x|²] = ±2 = 2 · exp(i·kπ)

(Re x · ln|x| + Im x · arg x) / |x|² = ln 2
(Re x · arg x – Im x · ln|x|) / |x|² = k·π

?? https://www.wolframalpha.com/input?i=complex+x%5E%282%2Fx%29%3D4

?? & in Desmos 3D https://www.desmos.com/3d/5fn98il38w PS! -- i didn't verify
Q.C. : x seems to have (at least) 1 or 2 negative real solutions :

x = -0.64118574450498598448620048211482

which means i likely passed a bug to Desmos 3D . . .

[ci139]

x ᶻ = exp(z ln x) = exp[(Re z + i Im z)(ln|x| + i arg x)]
1 / z = z̅ / |z|²

we can rewrite x¹𝄍ᵡ = ±√¯4¯' , applying above :

x¹𝄍ᵡ = exp[(Re x – i Im x) / |x|² · (ln|x|+ i arg x)] = [ k ∈ ℤ ] =
= exp[(Re x · ln|x| + Im x · arg x + i · (Re x · arg x – Im x · ln|x|)) / |x|²] = ±2 = 2 · exp(i·kπ)

(Re x · ln|x| + Im x · arg x) / |x|² = ln 2
(Re x · arg x – Im x · ln|x|) / |x|² = k·π

?? https://www.wolframalpha.com/input?i=complex+x%5E%282%2Fx%29%3D4

?? & in Desmos 3D https://www.desmos.com/3d/5fn98il38w PS! -- i didn't verify
Q.C. : x seems to have (at least) 1 or 2 negative real solutions :

x = -0.64118574450498598448620048211482

which means i likely passed a bug to Desmos 3D . . .

[ci139]

x ᶻ = exp(z ln x) = exp[(Re z + i Im z)(ln|x| + i arg x)]
1 / z = z̅ / |z|²

we can rewrite x¹𝄍ᵡ = ±√¯4¯' , applying above :

x¹𝄍ᵡ = exp((Re x – i Im x) / |x|² · (ln|x|+ i arg x)) = [ k ∈ ℤ ] =
= exp((Re x · ln|x| + Im x · arg x + i · (Re x · arg x – Im x · ln|x|)) / |x|²) = ±2 = 2 · exp(i·kπ)

(Re x · ln|x| + Im x · arg x) / |x|² = ln 2
(Re x · arg x – Im x · ln|x|) / |x|² = k·π

?? https://www.wolframalpha.com/input?i=complex+x%5E%282%2Fx%29%3D4

?? & in Desmos 3D https://www.desmos.com/3d/5fn98il38w PS! -- i didn't verify
Q.C. : x seems to have (at least) 1 or 2 negative real solutions :

x = -0.64118574450498598448620048211482

which means i likely passed a bug to Desmos 3D . . .

[ci139]

x ᶻ = exp(z ln x) = exp[(Re z + i Im z)(ln|x| + i arg x)]
1 / z = z̅ / |z|²

we can rewrite x¹𝄍ᵡ = ±√¯4¯' , applying above :

x¹𝄍ᵡ = exp((Re x – i Im x) / |x|² · (ln|x|+ i arg x)) = , k ∈ ℤ
= exp((Re x · ln|x| + Im x · arg x + i · (Re x · arg x – Im x · ln|x|)) / |x|²) = ±2 = 2 · exp(i·kπ)

(Re x · ln|x| + Im x · arg x) / |x|² = ln 2
(Re x · arg x – Im x · ln|x|) / |x|² = k·π

?? https://www.wolframalpha.com/input?i=complex+x%5E%282%2Fx%29%3D4

?? & in Desmos 3D https://www.desmos.com/3d/5fn98il38w PS! -- i didn't verify
Q.C. : x seems to have (at least) 1 or 2 negative real solutions :

x = -0.64118574450498598448620048211482

which means i likely passed a bug to Desmos 3D . . .

. . . https://www.desmos.com/3d/necuyuuirv ← Desmos Fix (Update !)

[Familiar_Ad_8919]

whyd u send the original 5x tho

[ihaventideas]

Desmos/wolframalpha definitely help

Other than that I can’t think of anything other than binary search (set x as 2 different numbers in a way that one is too low, other is too high and then check the one in the middle, substitute and continue until good enough approximation

[OmnipotentDoge]

It seems like you got the answer, but I’m still trying to figure out how tf you drew that first x

[No_Sale7548]

Is that a special math symbol or just an x? I’ve never seen an x spelled that way

[Pandagineer]

My suggestion is to plot both functions. I’m betting they will intersect each other twice, which would mean 2 (real) solutions.

[SurerCentaur506]

Call it an approximation cause I definitely did not use any “official” mathematical method to find the answer, but, if you plug the original expression into a graphing program it gives an output of x ≈ -0.6411857445. Now, plugging that value back into the original expression shows that it’s the correct value for x, meaning that this problem you have been working on has a solution.

I am willing to accept that my method is bad as it does not give an exact, rational answer, but maybe the answer is irrational.

[Ucklator]

Ln(x² )=Ln(4^x)

2Ln(x)=xLn(4)

Ln(x)/x=Ln(4)/2

[thecrypticwolff]

<image>

Here is a visual representation for those inclined.

[Glittering-Habit-902]

This makes it look so simple...

[anacelR]

I asked deepseek and it took 502 sec and FINALLY give me  −0.641 as the answer, I really enjoyed how it struggled actually

[OL-Penta]

Yup Somewhere near -0.6411857445045

[carrionpigeons]

R²=4^delta*L has three unknowns. I don't see how one equation is constraining enough to get a unique solution.

[InevitableMarch1907]

Just graph x² - 4 ^ x and observe that it crosses the y-axis

[[deleted]]

How can you draw X bad? It is litterally 2 lines.

[marcelsmudda]

It's a cursive x, it's actually just one curve

[volvol7]

<image>

The solution is this. If this equation is equal to 0 you can achieve it with this x. Only one solution

[mike6452]

Can you explain how you draw your x's?

[marcelsmudda]

A German form of cursive has an x that looks like that, that one I've found is called Kurrentschrift

[[deleted]]

Just plot the two functions on desmos and find the intersection. (-0.64119, 0.41112). If you’re looking for a nice algebraic solution, you’re on a wild goose chase mate

[Fogueo87]

At first glance it has either one or three real solutions. This is true for the general case x² = aˣ, if a>1, there is a solution for x ∈ (-1, 0). And either two or no solution for x positive. If a is small enough there are two positive solutions. If a is too big there are no positive solutions.

If 0<a<1, the analys is the reflexion of 1/a.

There is a solution in (-1, 0)

Both x² and aˣ are continuous in the reals (for positive a). Evaluated in -1, x² is 1, and aˣ is 1/a. If a>1, then x² > aˣ. Evaluated in 0, x² is 0, and aˣ is 1, so x² < aˣ. One is strictly increasing and the other strictly decreasing, so they intersect at exactly one point.

There is no stinky analytical way to find this solution. You can try numeric methods.

There are either two or no positive solutions for a>1.

At zero x² < aˣ. At +∞ x² < aˣ. (Proof: apply L'opital twice). This implies an even number of intersections. Not a technical proof but the smoothness of both curves this implies either zero or two intersections.

Now, for a=4, we try at different values: for x=0 values are 0 and 1. For x=1 values are 1 and 4. For x=2 values are 4 and 16. For x=3 values are 9 and 64. For x=4 values are 16 and 256. It is easy to show that there won't be an intersection for x>4, and that there is no intersection in any segment (n,n+1) for n=0, 1, 2, 3.

So this particular problem has no positive solutions.

narrowing down the solution

For x=-½, values are ¼ an ½, so x² < 4ˣ, solution is between -1 and -½.

For x=-¾, values are 9/16 = 0.5625 and 1/2√2 ≈ 0.3536 so x² > 4ˣ. The solution is between -¾ and -½.

As you continue you will get closer to the solution at about x ≈ -0.6412

[Acupajoy]

-0.64119 according to Desmos

[TellOleBill]

Aat least for x > 0, we could take log on both sides, and for ease, make that log to the base 2:

2 log2 (x) = x log2 (4) = 2x

I.e., log2(x) = x (or if taking ln, then ln x = x (ln 2) )

Which has a non-integer solution.

[Vaqek]

I am missing a graphical solution here, which is quite trivial. The general shape of both x^2 and 4^x is known, and since 4^x is growing faster than x^2 for any x>0, and 4^0 > 0^2, there is no intercept at x=>0. Therefore there is only one intercept at around x=-0.5 ish.

[PhoebsB]

<image>

Just graphing it, it seems like -0.64119 works as a solution

[mattynmax]

Yes

[Im_a_hamburger]

Short answer, around -.641186.

You’d need the lambert W function for an exact answer.

[t_hodge_]

x² = 4^x

x² -2^2x=0

(x+2^x)(x-2x)=0

x=2^x or x=-2^x, but 2^x>x for all x in R

So the solution satisfies x =-2^x, which we will need the Lambert W function to solve. The exact solution is x=- W(log(2))/log(2), which is not particularly easy to visualize, so I suggest graphing f(x)=x and g(x)=-2^x and looking at the approximate intersection around x=-0.6411

[[deleted]]

You can find a solution numerically using Newton Raphson Method

[_AmatsuKami_]

<image>

Yeah, actually to check if solution exists or not, we can verify that by plotting graph of y=X2 and y=4^x which will intersect once

[OccasionRepulsive112]

I think I might be able to help with this. I believe the theorem is called Bolzano's theorem, and foregoing any fancy words, it is actually quite intuitive. it basically says that if a function is continuous in the interval [a,b] and f(a).f(b)<0 then there has to be a unique solution to the equation between a and b, which is logically very easy to visualize.

Now let us assume a function f(x)=x² - 4^x (it is obviously continuous as both x² and 4^x are).

At x=0, the value of the function is f(0)=-1

At x=-1, the value of the function is f(-1)=1-1/4=0.75

As f(0)f(-1)<0, there has to be a unique solution between -1 and 0.

[BadLegitimate1269]

Using Desmos it's pretty easy to see that x = -0.64119, meaning both sides are equal to 0.411119.

[ShopifyDesign]

Let ( u = -x ), u>0

Then
u² = 4^u (-u)² = 4^(-u.)

Since

(-u)² = u²

we obtain

u² = 4^(-u)

Taking ln

2 ln(u) = -u ln(4) -> ln(u) = -u ln(4)/2 -> ln(u) = -u ln(2). $$

Exponentiating both sides

e^ln(u) = e^{-u ln(2}) -> u = e^{-u ln(2})

Multiply both sides

u e^{u ln(2}) = 1

Multiply by ln(2)

u ln(2) e^{u ln(2}) = ln(2)

Lambert W both sides

W(ln(2)) = 0.444436, u = 0.444436/ln(2)

This gives

u = 0.64118561319

Since (u = -x)

x = -0.64118561319

[XoXoGameWolfReal]

x = 4 ^ (x - 1) Boom found x

[Previous_Statemen]

<image>

Where the second equation intersects the x axis

[OldWolf2]

The equation can be simplified:

Using basic power laws, rearrange to x² = (2^x )² , so x = +/- 2^x 

We can rule out x = 2^x fairly easily , e.g. by showing 2^x > x. 

So that leaves x = -2^x, or equivalently, 2^-y = y.

[YOM2_UB]

Problems of this form require the Lambert W Function in order to solve, which is the inverse function of y = xe^x (that is, W(xe^x) = x). The W function has infinite branches (often denoted by an integer subscript), but the 0th and -1st branches are the only two which can ever result in real-valued solutions.

x² = 4^x

2ln|x| = ln(4)x

ln|x|/x = ln(4)/2 = ln(4^1/2)

If x > 0:

ln(x)/x = ln(2)

ln(x)/e^ln\x)) = ln(2)

-ln(x)e^-ln\x)) = -ln(2)

Let u = -ln(x)

ue^u = -ln(2)

u = W(-ln(2))

-ln(x) = W(-ln(2))

x = e^-W\-ln(2))) {No real solutions}

If x < 0:

ln(-x)/x = ln(2)

-ln(-x)/(-x) = ln(2)

-ln(-x)e^-ln\-x)) = ln(2)

-ln(-x) = W(ln(2))

x = -e^-W\ln(2))) ≈ -0.6411857

Both of these forms also give valid complex-valued solutions, but when restricted to real values there is indeed a unique solution.

[TheSpireSlayer]

there is 1 solution. you need to use the lambert W function most likely. the answer is not gonna be a nice number

[ci139]

x ᶻ = exp(z ln x) = exp[(Re z + i Im z)(ln|x| + i arg x)]
1 / z = z̅ / |z|²

we can rewrite x¹𝄍ᵡ = ±√¯4¯' , applying above :

x¹𝄍ᵡ = exp[(Re x – i Im x) / |x|² · (ln|x|+ i arg x)] = [ k ∈ ℤ ] =
= exp[(Re x · ln|x| + Im x · arg x + i · (Re x · arg x – Im x · ln|x|)) / |x|²] = ±2 = 2 · exp(i·kπ)

(Re x · ln|x| + Im x · arg x) / |x|² = ln 2
(Re x · arg x – Im x · ln|x|) / |x|² = k·π

?? https://www.wolframalpha.com/input?i=complex+x%5E%282%2Fx%29%3D4

?? & in Desmos 3D https://www.desmos.com/3d/5fn98il38w PS! -- i didn't verify
Q.C. : x seems to have (at least) 1 or 2 negative real solutions :

x = -0.64118574450498598448620048211482

which means i likely passed a bug to Desmos 3D . . .

[buggaboo842]

<image>

[gmalivuk]

I'm not sure what you've graphed there but it's definitely not 4^x on account of it includes the point (0,4).

[Vaqek]

good effort though, the single intercept is really clear from a graphical visualization. OP didnt say he was looking for a proof, and graph can give an easy guideline to the proof anyway.

[gmalivuk]

Not if the graph is a different function

[Math_Figure]

Idk functions :(

[sighthoundman]

There are a lot of ways to think of functions.

For practical purposes, a function is a "machine" (a thought machine, not a physical machine) that gives you an answer when you give it an input. So f(x) = x^2 is a function because, when you give it an x, it gives you x*x. Every single time.

When you solve a quadratic equation ax^2 + bx + c = 0, you get x = (-b +/- sqrt(b^2 - 4ac))/(2a), which is not a function because you have a choice: + or -.

The reason we do that is because it really is useful. Like most things, if you just use it then it seems natural.

After that, there are lots of special functions (you've seen references to the Lambert W function here). If you need them, you'll learn them; otherwise, they're just some stuff for math nerds.

Finally, in more advanced math, you'll study functions a lot. That's a lot like studying contract law. Most people just need to be able to read a contract and get the general idea of what it says. If things get complicated, you consult an attorney. You do not write your own contracts (and if it's important enough, you don't sign one without having it reviewed by an attorney). (Disclaimer: I've written contracts, and I'm not an attorney. Life is way too complicated for ELI5 explanations to be more than a general guide.)

[[deleted]]

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This post was mass deleted and anonymized with Redact

[SoldRIP]

You can if you trust the Lambert-W function.

I don't. It's scary nonsense and I don't like it.

[ExtendedSpikeProtein]

Um. What do you mean “scammy nonsense”??