# Natural Logarithms

1. Mar 7, 2009

### Compaq

ln(x+1) + ln(x+3) < ln(x+7)

Find x!

I should of course know this, second year at high school (in Norway). I've just forgotten the way attack the whole thing^^

2. Mar 7, 2009

### HallsofIvy

Staff Emeritus
ln(x+1)+ ln(x+3)= ln((x+1)(x+3)

To find x such that ln((x+1)(x+3))< ln(x+7), it is probably simplest to first solve
ln((x+1)(x+3))= ln(x+7) which, because ln is a "one-to-one" function, is equivalent to (x+1)(x+ 3)= x+7, a quadratic equation. The solutions to that equation separate the number line into intervals where ln(x+ 1)+ ln(x+3)< ln(x+7 for every point in the interval or is not true for every point.

3. Mar 7, 2009

### Compaq

Ahh, I see. By factorising with one of the logarithm rules, I get the ln to one number equals the ln to another number, thus the numbers must be the same.

Thanks, this + other tasks are to be handed in on Monday:)

PS, not sure if this was put in the correct forum.

So, anyway. Those of you who answer our questions. Who, or what, are you? Students? Educated people just coming in here regularly to answer?

Regards, Compaq

4. Mar 7, 2009

### Compaq

I got:

x = -14 V x = 11

Is that correct?

5. Mar 7, 2009

### Mentallic

Different people, different reasons for being here I often need help since some of my teachers don't know the answers to my problems (yes, even they are human), but I feel I need to give back to the community, so I help out where I can. Of course, ultimately we are all here to learn a thing or two.

Um not quite.

where you left off: $$(x+1)(x+3)<x+7$$

expanding and collecting like terms: $$x^2+3x-4<0$$

factorising: $$(x+4)(x-1)<0$$

Now from here, I find it easiest to graph the quadratic $$y=(x+4)(x-1)$$ and since y<0, the domain is for all values of x which y<0 (looking at the graph, wherever the quadratic is below the x-axis).

If you could follow that, you should find that $$(x+4)(x-1)<0$$, then $$-4<x<1$$

ok so it looks like it is answered, but because we manipulated the problem in such a way as to rid ourselves of the logarithm, there might be more restrictions on the value of x. Look back at the original question, for $$ln(x)$$ what can't the value of 'x' be?

6. Mar 8, 2009

### Compaq

Yeah, I screwd up with the general forumula for quadrics (or whatever you call itXD), so I tried with that other way (which I don't know the English term of).. (I add (b/2)2 and then add -(b/2)2 ) I got the same factors as you, at least.

And we're dealing with the logarithm function, so x > 0, negative numbers, nor sero, have logarithms, logically. So:

(x+4)(x-1) < 0 when 0<x<1

Does that seem right, then:)

And thanks a lot:D

One more question, how do you add fractions and all that stuff, I feel I'm using a werid method here:P

7. Mar 8, 2009

### Compaq

Ohh, and I always use factor tables...

8. Mar 8, 2009

### Compaq

Wait! Of course it should be:

ln(x+1) + ln(x+3) < ln(x+7) when 0<x<1

9. Mar 9, 2009

### Mentallic

Sure, do it however you like. Factoring, using the quadratic formula and completing the square all work.

Not exactly. For $$ln(a), a>0$$
Which means in the question, $$ln(x+1) + ln(x+3) < ln(x+7)$$

For each term inside the logarithm: $$x+1>0 , x+3>0 , x+7>0$$

Simplifying: $$x>-1 , x>-3 , x>-7$$

But all these inequalities of x must be satisfied, so take the largest: $$x>-1$$

So what is the final result then? Remember: both the answer we got from finding the inequality $$(x+4)(x-1)<0$$ and $$x>-1$$ must be satisfied.