# Inequality with Logarithms

1. Aug 23, 2015

### Cosmophile

1. The problem statement, all variables and given/known data
$$\ln \frac{(x+1)}{(x-1)} \geq 0$$

2. Relevant equations
$$\ln \frac {a}{b} = \ln a - \ln b$$
$$\ln \frac {(x+1)}{(x-1)} = \ln (x+1) - \ln (x-1)$$

3. The attempt at a solution
$$\ln (x+1) \geq \ln (x-1)$$
$$e^{\ln (x+1)} \geq e^{\ln (x-1)}$$
$$(x+1) \geq (x-1)$$

According to Wolfram, the solution is $x > 1$, but my skills with logs and exponentials are clearly very rusty.

2. Aug 24, 2015

### Staff: Mentor

This seems reasonable, but unfortunately, isn't valid.

Your original inequality is equivalent to $\frac{x + 1}{x - 1} \ge 1$, which you can get in one step. If the log of something is >= 0, then the something has to be >= 1.
If you multiply both sides by x - 1, you get what you show in the last line, above. The problem is that you don't know whether x - 1 is positive or negative (or zero), so it's impossible to say whether the direction of the inequality stays the same or is reversed.

To get around this problem, multiply instead by $(x - 1)^2$. You should also explicitly note what values of x are allowed.

3. Aug 24, 2015

### Cosmophile

I see how you got $\frac {x+1}{x-1} \geq 1$. Why am I okay to multiply by $(x-1)^2$, and why is this necessary? Thank you for the help as always!

Also, "if the log of something is > 0, then the something has to be >1" Why? This is where my understanding of logarithms (or lack thereof) is on full display.

4. Aug 24, 2015

### Staff: Mentor

You can multiply both sides of an inequality by any positive number, without changing the direction of the inequality. $(x - 1)^2$ is positive, provided that $x \neq 1$.
$\ln a \ge 0$
$\Leftrightarrow e^{\ln a} \ge e^0 = 1$
$\Leftrightarrow a \ge 1$

Of course, in the above, a must be positive.

5. Aug 24, 2015

### Cosmophile

So, are you multiplying both sides of the inequality by $(x-1)^2$ then?

6. Aug 24, 2015

### Staff: Mentor

Yes. Multiply both sides of $\frac{x + 1}{x - 1} \ge 1$ by $(x - 1)^2$. Be sure to specify any restrictions on possible values of x, keeping in mind what the domain of the ln() function is (which the original inequality involves).

7. Aug 24, 2015

### MindWalk

In your attempt at a solution, you get ln (x + 1) >= ln (x - 1). From here you need to think about two things: (1) When is the ln function defined? (2) When is ln (a bigger number) >= ln (a smaller number)? Thinking about these should get you the solution.

8. Aug 24, 2015

### Ray Vickson

Look at the plot of the log function y = ln (z) for z > 0. You will see that for 0 < z < 1 we have ln z < 0, but for z > 1 we have ln z > 0.

Anyway, you want $(x+1)/(x-1) \geq 1$. You do not need to multiply by $(x-1)^2$; that is not the only way to do it. An alternative is to recognize that we need the numerator and denominator to have the same sign, so there are two cases.
(1) If the numerator and denominator are both > 0 the inequality equivalent to $x+1 > x-1$; but we also have $x-1 > 0$. Altogether, this just says that $x > 1$.
(2) If the numerator and denominator are both < 0 the inequality is equivalent to $x+1 \leq x-1$, with $x < -1$ (note the reversal from '>' to '<'). However, this is clearly impossible., so we cannot obey the desired inequality when $x < -1$.

This is not so much a problem in logarithm manipulation as it is a problem in inequality manipulation.

9. Aug 24, 2015

### pasmith

Alternatively, $$\frac{x+1}{x-1} = \frac{(x - 1) + 2}{x-1} = 1 + \frac{2}{x-1}.$$ Thus, since $2 > 0$, we have $\frac{x+1}{x-1} \geq 1$ when $x - 1 > 0$.

10. Aug 25, 2015

### Cosmophile

It's obvious at this point that I desperately need to work on my conceptual understanding of logs, exponentials, and inequalities. Do any of you have recommendations on places to start? I'm reading through Lang's calculus text, and I know he has a section on logs and exponentials. He starts the book with inequalities, but it apparently doesn't cover all that I need to know on the topic.

Should I be concerned that a multiplication by the square of the denominator was not obvious, or is that simply a trick you learned along the way? Also, the only restriction I see is that $x > 1$, as $\ln x$ is defined $\forall x \in \mathbb{R} | x >0$. Is this all I needed to observe from the beginning? Thanks for the assistance so far!

(1) $\ln(x)$ is defined $\forall x \in \mathbb{R} | x>0$.
(2) $\ln(n) > \ln(m)$ is always true for $n > m$, is it not? $\ln(x)$ is always increasing.

Right, I see now why the thing I'm taking the log of must be >= 1 if the log is > 0. Now, for your first case, you say:

"(1) If the numerator and denominator are both > 0 the inequality equivalent to $x+1 > x-1$; but we also have $x-1 > 0$. Altogether, this just says that $x > 1$."

How did you gather that $x+1 > x-1$ and $x-1 > 0$ together say $x > 1$? This is another great example of my blatant ignorance and inexperience with inequalities, so I appreciate the assistance. Of course, I can see immediately from $x -1 > 0$ that $x>1$; I'm simply confused as to how you get the same result from $x + 1> x-1$. This inequality holds true for all x (which is why the second case is clearly impossible if I'm not mistaken).

Oh, this is a pretty neat way to approach it. This works because $1 + \frac {2}{x-1} > 0 \quad \forall x > 1$, correct?

11. Aug 25, 2015

### Qwertywerty

What you need to remember while solving inequalities is, simply this - unless you are sure that a particular function is always positive, or negative, for that matter, you can not just simply cross multiply across an inequality.
For example, if you had ( x+1 )/( x2+1 ) <= 1, you could cross multiply ( Why ? ).

An alternative solution would be to simply bring all quantities to one side of the inequality, and taking LCM, form the intervals for which that inequality is positive and negative.

Hope this helps.

12. Aug 25, 2015

### Ray Vickson

13. Aug 25, 2015

### SammyS

Staff Emeritus
Ray,

You didn't get a "normal" font, because you replied inside of the quote of Cosmophile .

14. Aug 25, 2015

### Ray Vickson

AFAIK, I always do that and have never before had the problem!