# Analytic Solution to ##x^{\alpha} +x =1##?

• I
• bob012345
In summary: That will require ##x>1##. And I assume you want to solve for ##x##, not for...?Yes, I was hoping to find a solution for ##x## in terms of ##\alpha##.Yes, I was hoping to find a solution for ##x## in terms of ##\alpha##.

#### bob012345

Gold Member
TL;DR Summary
Is there an analytic solution to the simple equation ##x^{\alpha} +x =1## where ##\alpha## is a constant ?
Is there an analytic solution to the simple equation ##x^{\alpha} +x =1##? I can get to a solution by iteration or by graphical methods but I wish to find a closed form exact solution. ##\alpha## is a constant. I tried to put it into a form where I could use the quadratic formula but that didn't work i.e something like ##c^{2\beta} + c^{\beta} =1##. Thanks.

Delta2
bob012345 said:
Summary: Is there an analytic solution to the simple equation ##x^{\alpha} +x =1## where ##\alpha## is a constant ?

Is there an analytic solution to the simple equation ##x^{\alpha} +x =1##? I can get to a solution by iteration or by graphical methods but I wish to find a closed form exact solution. ##\alpha## is a constant. I tried to put it into a form where I could use the quadratic formula but that didn't work i.e something like ##c^{2\beta} + c^{\beta} =1##. Thanks.
There is generally no analytic solution. E.g. consider the case ##\alpha=7## then ##x^7+x-1=0## cannot be solved analytically.

dextercioby, Delta2 and bob012345
fresh_42 said:
There is generally no analytic solution. E.g. consider the case ##\alpha=7## then ##x^7+x-1=0## cannot be solved analytically.
Thanks. I suppose it is the same if ##\alpha## is less than 2 with the obvious exceptions of 0, and 1? My interest in this question is non-integer exponents.

bob012345 said:
Thanks. I suppose it is the same if ##\alpha## is less than 2 with the obvious exceptions of 0, 1 and 2?
There is a theorem that says that there are no solutions with radicals (##\sqrt[n]{.}##) for arbitrary integer polynomials of degree ##5## or higher. We can solve it for ## \alpha \in \{0,1,2,3,4\}## by roots. So ##\alpha \geq 5## has good chances of not being solvable. And what about non-integer values?

fresh_42 said:
There is a theorem that says that there are no solutions with radicals (##\sqrt[n]{.}##) for arbitrary integer polynomials of degree ##5## or higher. We can solve it for ## \alpha \in \{0,1,2,3,4\}## by roots. So ##\alpha \geq 5## has good chances of not being solvable. And what about non-integer values?
Well, turning the problem around it is easy to solve for ##\alpha## given any ##x##.

$$\large x^{\alpha} + x=1$$

$$\alpha log(x) = log (1-x)$$

$$\alpha = \large \frac{log(1-x)}{log(x)}$$

I just thought one might be able to transform the equation around to solve for ##x## given an ##\alpha## for this specific equation form by some kind of substitution.

Delta2
bob012345 said:
Well, turning the problem around it is easy to solve for ##\alpha## given any ##x##.

$$\large x^{\alpha} + x=1$$

$$\alpha log(x) = log (1-x)$$

$$\alpha = \large \frac{log(1-x)}{log(x)}$$

I just thought one might be able to transform the equation around to solve for ##x## given an ##\alpha## for this specific equation form.
If you put it that way, we are given an equation with only one occurrence of the variable, so that we only have ##c_{1}^\alpha = c_{2}## which is basically the definition of the logarithm.

The problem with ##x^\alpha +x-1=0## when we ask for ##x## is, that we have a mixture of some arbitrary multiplication and a few simple additions. Whenever multiplication and addition meet, things get complicated. There is actually only one single law that combines them:
$$a\cdot (b+c) = a\cdot b + a \cdot c$$
That's it. It is all we have. And every combination of the two that differs from the distributive law is a complex problem, in our case literally complex.

Delta2 and bob012345
All alpha values where I see a solution:

##\alpha \in \{-3,-2,-1,0,1,2,3,4\}## as we get a solvable polynomial, multiply by powers of x for negative ##\alpha##.
##\alpha \in \{-\frac 1 3, -\frac 1 2, \frac 1 2, \frac 1 3, \frac 1 4\}## as we get a solvable polynomial in the 2nd, 3rd, 4th root of x.
For ##\alpha=5+6z## for integer z, ##e^{\pm \pi i/3}## is a solution. There is no general closed solution for polynomials of that degree, but some special cases can still have closed solutions.

dextercioby, bob012345 and Delta2
Where this was all coming from is from a similar equation used to derive it
$$\large a^y + b^y =1$$ where ##a,b## are constants and I let ##\large x= b^y## then we can write ##\large a^y = b^{\alpha y} = (b^y)^{\alpha} = x^{\alpha}## giving $$\large x^{\alpha} + x=1$$

I suppose the first form is just as problematic.

bob012345 said:
I suppose the first form is just as problematic.
Yes. Can you at least say where ##a,b,y## are taken from?

fresh_42 said:
Yes. Can you at least say where ##a,b,y## are taken from?
Yes, there was a problem but in its original form used ##x## which goes as $$16^x + 20^x =25^x$$ solve for ##x##. I divided both sides by ##25^x##.

I solved that exactly analytically but then wanted to generalize the coefficients a bit.

Last edited:
\begin{align*}
16^x+20^x=25^x &\Longleftrightarrow (5^x-4^x)(5^x+4^x)=4^x\cdot 5^x \\
&\Longleftrightarrow \left(\left(\dfrac{1}{4}\right)^x-\left(\dfrac{1}{5}\right)^x\right)
\left(\left(\dfrac{1}{4}\right)^x+\left(\dfrac{1}{5}\right)^x\right) =1\\
&\Longleftrightarrow \left(\dfrac{1}{4^x}\right)^2+1^2=\left(\dfrac{1}{5^x}\right)^2
\end{align*}
is a very specific form: ##(a^x)^2+1^2=(b^x)^2.## Now we have a quadratic equation and we can e.g. look for Pythagorean triples that are all known.

As a rule of thumb, you can say that, whenever there are symmetries, then there is a good chance to solve a system.

dextercioby and bob012345
fresh_42 said:
There is a theorem that says that there are no solutions with radicals (##\sqrt[n]{.}##) for arbitrary integer polynomials of degree ##5## or higher. We can solve it for ## \alpha \in \{0,1,2,3,4\}## by roots. So ##\alpha \geq 5## has good chances of not being solvable. And what about non-integer values?
Is that no Real solutions? Is ##\alpha## Real? How about Complex ones? Seems like something like the W function may work?

WWGD said:
Is that no Real solutions? How about Complex ones? Seems like something like the W function may work?
This is irrelevant. ##p(x)\in \mathbb{Z}[x]## is the restriction.

fresh_42 said:
This is irrelevant. ##p(x)\in \mathbb{Z}[x]## is the restriction.
But ##Z[x]## refers to polynomials, doesn't it? So we assume ##\alpha \in \mathbb Z##?

bob012345 said:
Well, turning the problem around it is easy to solve for ##\alpha## given any ##x##.

$$\large x^{\alpha} + x=1$$

$$\alpha log(x) = log (1-x)$$

$$\alpha = \large \frac{log(1-x)}{log(x)}$$

I just thought one might be able to transform the equation around to solve for ##x## given an ##\alpha## for this specific equation form by some kind of substitution.
That will require ##x>1##. And I assume you want to solve for ##x##, not for ##\alpha##.

You mean for the problem here? No, we didn't assume anything. Integers are simply a domain we know something about.

fresh_42 said:
\begin{align*}
16^x+20^x=25^x &\Longleftrightarrow (5^x-4^x)(5^x+4^x)=4^x\cdot 5^x \\
&\Longleftrightarrow \left(\left(\dfrac{1}{4}\right)^x-\left(\dfrac{1}{5}\right)^x\right)
\left(\left(\dfrac{1}{4}\right)^x+\left(\dfrac{1}{5}\right)^x\right) =1\\
&\Longleftrightarrow \left(\dfrac{1}{4^x}\right)^2+1^2=\left(\dfrac{1}{5^x}\right)^2
\end{align*}
is a very specific form: ##(a^x)^2+1^2=(b^x)^2.## Now we have a quadratic equation and we can e.g. look for Pythagorean triples that are all known.

As a rule of thumb, you can say that, whenever there are symmetries, then there is a good chance to solve a system.
Yes, this was a special case since it could be reduced to a quadratic because of the specific values of the coefficients. BTW, the way I did it was divide by the RHS, combine to get

$$(16/25)^x + (20/25)^x =1$$ which is

$$(0.64)^x + (0.8)^x =1$$

which is $$(0.8^x)^2 + (0.8)^x =1$$ Then substituted ## z=0.8^x##

$$z^2 + z =1$$ which gives ##z= \frac{-1±5^{1/2}}{2}##

and ## x=\large \frac{log(z)}{log(0.8)}##Then, I wanted to see if I changed the 16 to say, 18, if it had an analytic solution. The answer is no in general.

$$18^x+20^x=25^x$$

fresh_42
BTW, ##x= -∞## might work too for ##a^x + b^x = c^x##.

bob012345 said:
BTW, ##x= -∞## might work too for ##a^x + b^x = c^x##.
As does ##x=+\infty ## but that should only be used if we explicitly state that hyperreal numbers are allowed. In any other, i.e. standard case, there is no number ##\pm \infty ## and we cannot pretend there was.

bob012345 said:
BTW, ##x= -∞## might work too for ##a^x + b^x = c^x##.
In the sense that for ##a, b, c > 1##:$$\lim_{x \to -\infty} (a^x + b^x) = \lim_{x \to -\infty} (c^x) = 0$$

bob012345

## 1. What is an analytic solution?

An analytic solution is a mathematical method for finding the exact solution to a problem, as opposed to a numerical solution which uses approximations. It involves using algebraic operations and equations to find the solution.

## 2. How do you solve ##x^{\alpha} +x =1## analytically?

To solve this equation analytically, we first need to isolate the variable with the exponent. This can be done by subtracting x from both sides. Then we can use logarithms to get rid of the exponent, and finally solve for x using algebraic operations.

## 3. What is the significance of the ##\alpha## in the equation?

The ##\alpha## is a variable that represents the power to which x is raised. It allows us to solve for a range of equations with different exponents, rather than just a specific one.

## 4. Are there any limitations to using an analytic solution?

Yes, there are limitations to using an analytic solution. It may not be possible to find an exact solution for all problems, and some equations may require complex mathematical operations that are difficult to solve.

## 5. Can an analytic solution be used for real-world problems?

Yes, analytic solutions can be used for real-world problems in various fields such as physics, engineering, and economics. However, it may not always be the most practical or efficient method, and in some cases, numerical solutions may be preferred.