Solve (x + 1)^(x - 1) = (x - 1)^(x + 1)

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Discussion Overview

The discussion revolves around solving the equation $$(x + 1)^{x - 1} = (x - 1)^{x + 1}$$ in the real numbers. Participants explore various approaches to the problem, including logarithmic manipulation, graphical analysis, and alternative algebraic methods.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a logarithmic approach to simplify the equation, leading to the conclusion that $m = x - 1$ can be used to reformulate the problem.
  • Another participant suggests analyzing the graph of the function to determine the behavior of the equation, noting that the derivative is negative for values greater than 3 and indicating no solutions exist between 0 and 3.
  • There is a discussion about the behavior of the function for negative values of $x$, with one participant proposing to consider cases where the bases are negative and how that affects the solutions.
  • Some participants identify $x = 3$ and $x = -3$ as solutions, while others express uncertainty about whether these are the only solutions.
  • A participant introduces an alternative method that avoids logarithms, suggesting a different approach to arrive at the same result, which is acknowledged as quicker and neater by another participant.

Areas of Agreement / Disagreement

Participants generally agree on the identified solutions of $x = 3$ and $x = -3$, but there is no consensus on whether these are the only solutions. Multiple competing views on the methods and interpretations of the problem remain present.

Contextual Notes

Participants note that the function is only defined for specific values of $x$, particularly for negative numbers, which complicates the analysis. There are also discussions about the need for careful consideration of the signs in the equations when dealing with negative bases.

Who May Find This Useful

This discussion may be useful for individuals interested in advanced algebraic techniques, graphical analysis of functions, and those exploring the nuances of solving equations involving exponentials and logarithms.

anemone
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Hi MHB,

Recently I came across this problem "Solve the equation in $R$ of $$(x+1)^{x-1}= (x-1)^{x+1}$$ and I only managed to solve it partially and here is what I have done:

$$(x+1)^{x-1}= (x-1)^{x+1}$$

$$\ln(x+1)^{x-1}=\ln(x-1)^{x+1}$$

$$(x-1)\ln(x+1)=(x+1)\ln(x-1)$$

$$x\ln(x+1)-\ln(x+1)=x\ln(x-1)+\ln(x-1)$$

$$x\ln \left(\frac{x+1}{x-1}\right)=\ln(x-1)(x+1)$$

And let $m=x-1$ we then have

$$(m+1)\ln \left(\frac{m+2}{m}\right)=\ln(m)(m+2)$$

$$m\ln \left(\frac{m+2}{m}\right)+\ln \frac{m+2}{m}=\ln(m)(m+2)$$

$$m\ln \left(\frac{m+2}{m}\right)=\ln \left((m)(m+2)\right)\left(\frac{m}{m+2}\right)$$

$$m\ln \left(\frac{m+2}{m}\right)=\ln m^2$$

$$\ln \left(\frac{m+2}{m}\right)^m=\ln m^2$$

By equating the argument from both sides, we obtain

$$\left(\frac{m+2}{m}\right)^m=m^2$$

The answer is then checked when $m=2$, i.e. $x=3$.

My question is, how do I determine that $-3$ is another answer to this problem and also, how do I make sure these are the only two answers to this problem?

Thanks in advance for any helps that anyone is going to pour in.
 
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Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

Hey anemone!

To prove there are no other solutions, we can analyze the graph and the derivative of $$(x+1)^{x-1} - (x-1)^{x+1}$$.
A look at the graph shows the derivative is negative for x values greater than 3.
The graph also shows there are no other solutions between 0 and 3.

That leaves the negative numbers that show funny behavior. The function is only defined for specific negative numbers, since it is supposed to be real.

To work around it we can bring the minus sign outside.
If the powers exist, they are either negative or positive.
That way we get a differentiable function again.

Note that:
$$(x+1)^{x-1} = \pm (-x-1)^{x-1}$$
$$(x-1)^{x+1} = \pm (-x+1)^{x+1}$$

First case (taking plus 2 times) is $$(-x-1)^{x-1} - (-x+1)^{x+1}$$.
Looking at the graph shows it has a solution at x = -3.
And looking at its derivative shows there will be no other solutions.
Second case (taking minus 2 times) gives the same result.
Third and fourth case (alternating signs) give no new results.

Edited: fixed typos as noted it following post.
 
Last edited:
Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

Thank you very much for the reply, I like Serena!:)

Hmm...I still don't quite understand...:(

I like Serena said:
To prove there are no other solutions, we can analyze the graph and the derivative of $$(x+1)^{x-1} - (x-1)^{x+1}$$.
A look at the graph shows the derivative is negative for x values greater than 3.
The graph also shows there are no other solutions between 0 and 3.

By this do you mean we've to first try to sketch its graph on our own?

I like Serena said:
Note that:
$$(x+1)^{x-1} = \pm (-x-1)^{x-1}$$ ---(1)
$$(x-1)^{x+1} = \pm (-x+1)^{x-1}$$---(2)

Hmm...I believe (2) is meant for $$(x-1)^{x+1} = \pm (-x+1)^{x+1}$$ and hence $$(-x-1)^{x-1} - (-x+1)^{x-1}$$ should be $$(-x-1)^{x-1} - (-x+1)^{x+1}$$, is that so, I like Serena?
 
Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

anemone said:
Thank you very much for the reply, I like Serena!

Sure! (Wink)
anemone said:
By this do you mean we've to first try to sketch its graph on our own?

Yes.
I would create a schematic with a number of values for the function and also for its derivative.
That is enough to create a rough graph that can chart zero crossings and extremes.
We can't directly find zeroes or zeroes of the derivative, but we can see where they switch sign.

If we can see that the derivative will be negative from x=3 and onward, and we also see that the function value is 0 at x=3, we can say for sure there will be no more zeroes beyond x=3.
Hmm...I believe (2) is meant for $$(x-1)^{x+1} = \pm (-x+1)^{x+1}$$ and hence $$(-x-1)^{x-1} - (-x+1)^{x-1}$$ should be $$(-x-1)^{x-1} - (-x+1)^{x+1}$$, is that so, I like Serena?

Yes. That is so.
How sloppy of me. (Tmi)
I have edited my previous post to fix it.
 
Last edited:
Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

Hello, anemone!

\text{Solve in }\mathbb{R}:\;(x+1)^{x-1}\,=\, (x-1)^{x+1}
"Eyeballing" the problem, we have: .a^b \,=\,b^a

The two obvious solutions are:
. . 4^2 \,=\,2^4 \;\text{ and }\;(\text{-}2)^{\text{-}4} \,=\,(\text{-}4)^{\text{-}2}

That is: .x \,=\, \pm3
 
Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

I like Serena said:
Yes.
I would create a schematic with a number of values for the function and also for its derivative.
That is enough to create a rough graph that can chart zero crossings and extremes.
We can't directly find zeroes or zeroes of the derivative, but we can see where they switch sign.

If we can see that the derivative will be negative from x=3 and onward, and we also see that the function value is 0 at x=3, we can say for sure there will be no more zeroes beyond x=3.

Thank you I like Serena for the further reply...yeah! I always love sketching graph without the help of graphical calculator and I'm going to try sketching this and see if I noticed everything as you mentioned and your help is very much appreciated!(Nerd)

I like Serena said:
Yes. That is so.
How sloppy of me. (Tmi)
I have edited my previous post to fix it.

Oh okay and thanks!

soroban said:
Hello, anemone!


"Eyeballing" the problem, we have: .a^b \,=\,b^a

The two obvious solutions are:
. . 4^2 \,=\,2^4 \;\text{ and }\;(\text{-}2)^{\text{-}4} \,=\,(\text{-}4)^{\text{-}2}

That is: .x \,=\, \pm3

Hi soroban,

Thank you so much!
 
Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

Also, Anemone, there's a slightly faster way to arrive at the same result you did, without logarithms:

$$(x+1)^{x-1}= (x-1)^{x+1}$$

$$\frac{(x+1)^{x-1}}{(x - 1)^2} = \frac{(x-1)^{x+1}}{(x - 1)^2} ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ x \ne 1$$

$$\frac{(x+1)^{x-1}}{(x - 1)^2} = (x-1)^{x-1}$$

$$\frac{(x+1)^{x-1}}{(x-1)^{x-1}} = (x - 1)^2$$

$$\left ( \frac{x+1}{x - 1} \right )^{x - 1} = (x - 1)^2$$

Now substitute $m = x - 1$ and we get:

$$\left ( \frac{m + 2}{m} \right )^m = m^2$$

I just thought it was worth sharing.
 
Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

Bacterius said:
Also, Anemone, there's a slightly faster way to arrive at the same result you did, without logarithms:

$$(x+1)^{x-1}= (x-1)^{x+1}$$

$$\frac{(x+1)^{x-1}}{(x - 1)^2} = \frac{(x-1)^{x+1}}{(x - 1)^2} ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ x \ne 1$$

$$\frac{(x+1)^{x-1}}{(x - 1)^2} = (x-1)^{x-1}$$

$$\frac{(x+1)^{x-1}}{(x-1)^{x-1}} = (x - 1)^2$$

$$\left ( \frac{x+1}{x - 1} \right )^{x - 1} = (x - 1)^2$$

Now substitute $m = x - 1$ and we get:

$$\left ( \frac{m + 2}{m} \right )^m = m^2$$

I just thought it was worth sharing.

Hey Bacterius,

I'm so glad that you shared this with me! Thanks!(Nerd)

And it's so obvious that your method is quicker and neater than mine! Bravo!
 

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