# Cute Algebra Problem Is This Correct?

1. Aug 5, 2008

1. The problem statement, all variables and given/known data
Solve for x
$$\left( \frac{x+1}{x} \right)^x \left( ln \left(\frac{x}{x+1} \right)+\frac{1}{x+1}\right)=0$$

2. Relevant equations
Is the solution correct?

3. The attempt at a solution
$$\frac{ \left( \frac{x+1}{x} \right)^x } {\left( \frac{x+1}{x} \right)^x} \left( ln \left(\frac{x}{x+1} \right)+\frac{1}{x+1}\right)=\frac{0}{\left( \frac{x+1}{x} \right)^x}$$
$$ln \left(\frac{x}{x+1} \right)+\frac{1}{x+1}\right)=0$$
$$ln\left(x\right)-ln\left(x+1\right)+\left(x+1\right)^{-1}=0$$
$$e^{ln\left(x\right)}-e^{ln\left(x+1\right)}+e^{\left(x+1\right)^{-1}}=e^{0}$$
$$x-(x+1)+e^{\left(x+1\right)^{-1}}=1$$
$$x-x-1+e^{\left(x+1\right)^{-1}}=1$$
$$e^{\left(x+1\right)^{-1}}=1+1$$
$$ln\left(e^{\left(x+1\right)^{-1}}\right)=ln\left(2\right)$$
$$\left(x+1\right)^{-1}=ln\left(2\right)$$
$$\frac{1}{x+1}=ln\left(2\right)$$
$$1=\left(x+1\right)ln\left(2\right)$$
$$\frac{1}{ln\left(2\right)}=x+1$$
$$x=\frac{1}{ln\left(2\right)}-1$$

2. Aug 5, 2008

### rock.freak667

$$e^{ln\left(x\right)}-e^{ln\left(x+1\right)}+e^{\left(x+1\right)^{-1}}=e^{0}$$

is wrong since, $e^0 \neq 0$ and elnx=x NOT lnx

Last edited: Aug 5, 2008
3. Aug 5, 2008

### Ygggdrasil

$$e^{ln\left(x\right)-ln\left(x+1\right)+\left(x+1\right)^{-1}} \neq e^{ln\left(x\right)}-e^{ln\left(x+1\right)}+e^{\left(x+1\right)^{-1}}$$

4. Aug 5, 2008

Ygggdrasil: So what you are saying is that I can't distribute the e to each of the terms in the equation and raise them as its power rather I raise each side as a power of e as an entire term.

5. Aug 5, 2008

ok I see what I can't do but do you have an ideas how this could be solved?

6. Aug 5, 2008

### snipez90

I don't think there are any solutions. The first factor yields no zero. From the second factor, I got it down to solving [1 + 1/x]^(x+1) = e. But graphically it looks like the left hand side approaches e but never equals it.

I think the reason for this is that e is transcendental and also it can be written as an infinite series.

Last edited: Aug 5, 2008
7. Aug 5, 2008

### kraghunath

can you take the derivative on both sides? i know this is supposed to be just algebra...but just wondering if thats allowed?

8. Aug 5, 2008

Yes, I came to the conclusion that the series diverges because of the "Alternating Series Test" see original problem looked like $$\displaystyle \sum^{\infty}_{k=1} {\left(-1\right)^{k+1} \left( \frac{k}{k+1}\right)^{k}}$$
and I knew it failed one of the conditions of the test but I wanted to see if I could use the 1st Integral Test to see if the series was decreasing and what is above is my sad attempt to trying to find a critical number.

9. Aug 5, 2008

### snipez90

Yes, you are allowed to take the derivative of both sides in an algebraic equation. There is a simple condition that has to be met though I think(besides the fact that both functions have to be continuous and differentiable). Actually one of my ideas was to use calculus to prove that there are no solutions (at least that's what I think).

I haven't really taken the derivative of both sides of an equation much unless it's implicit differentiation.

10. Aug 5, 2008

Woops I meant to say I was using the 1st Deriv. Test not 1st Integral Test.

11. Aug 5, 2008

### kraghunath

ok so i tried taking the derivative on both sides and ended up with x = -1/2, which doesnt work when plugged back in...because the ln term...ok i lied...i got -1 = 0? not x = -1/2

Last edited: Aug 5, 2008
12. Aug 8, 2008

### Tobias Funke

You can take derivatives of both sides if the two sides are equal as functions, not just for certain values. This isn't the case here. For example, to solve 2x=x^2, you wouldn't differentiate both sides and get 2=2x and deduce that x=1.

13. Aug 8, 2008

### Gib Z

Showing that derivatives are equal only shows the original functions must differ by a constant. Whether or not that constant is zero is another matter.

Remember that if ab=0, a=0 and/or b=0.

Looking at the first factor, it only yields x=-1, but that is not allowed as a division by zero error occurs with the second factor.

Looking at the second factor, we let u=1/(1+x), such that we wish to solve $\log_e ( 1-u) + u =0$.

It's first derivative, $1/(u-1) + 1$, has only one zero, at u=0, which we note is a root of the equation we wish to solve. We also note that the function is monotonically increasing for u< 0, and monotonically decreasing for u>0, making it clear that u=0 is where the absolute maximum occurs, ensuring there are no more roots. But since $u=1/(1+x)$ and that $1/(1+x) = 0$ has no solution, we can conclude the original equation has no roots.