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

#### anemone

##### MHB POTW Director
Staff member
Hi MHB,

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

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

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

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

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

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

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

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

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

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

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

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

By equating the argument from both sides, we obtain

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

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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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 $$\displaystyle (x+1)^{x-1} - (x-1)^{x+1}$$.
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:
$$\displaystyle (x+1)^{x-1} = \pm (-x-1)^{x-1}$$
$$\displaystyle (x-1)^{x+1} = \pm (-x+1)^{x+1}$$

First case (taking plus 2 times) is $$\displaystyle (-x-1)^{x-1} - (-x+1)^{x+1}$$.
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.

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#### anemone

##### MHB POTW Director
Staff member
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...

To prove there are no other solutions, we can analyze the graph and the derivative of $$\displaystyle (x+1)^{x-1} - (x-1)^{x+1}$$.
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?

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

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

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 $$\displaystyle (x-1)^{x+1} = \pm (-x+1)^{x+1}$$ and hence $$\displaystyle (-x-1)^{x-1} - (-x+1)^{x-1}$$ should be $$\displaystyle (-x-1)^{x-1} - (-x+1)^{x+1}$$, is that so, I like Serena?
Yes. That is so.
How sloppy of me.
I have edited my previous post to fix it.

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#### soroban

##### Well-known member
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: .$$x \,=\, \pm3$$

#### anemone

##### MHB POTW Director
Staff member
Re: Solve (x+1)^(x-1)=(x-1)^(x+1)

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!

Yes. That is so.
How sloppy of me.
I have edited my previous post to fix it.
Oh okay and thanks!

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!

#### Bacterius

##### Well-known member
MHB Math Helper
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.

#### anemone

##### MHB POTW Director
Staff member
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.
Hey Bacterius,

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

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