# An interesting equation.

1. Sep 28, 2005

### Robokapp

can anyone explain the behaviour of this function? i looked at the graph and everything, but if i solve it one way i get a different answer than if i solve it another way.

Ln[(x+1)^2]=2

one way to solve it (What i did)

e^{Ln[(x+1)^2]}=e^2

(x+1)^2=e^2

(x+1)^2-e^2=0

(x+1-e)(x+1+e)=0

x=e-1 and x=-e-1

another way, what my math teacher did

Ln[(x+1)^2]=2

2Ln(x+1)=2

Ln(x+1)=1

X+1=e^1

x=e-1

as you can see somehow one of his roots, the -e-1 is just...gone. i checked the math 1 million times. this is puzzling me for a few months now...where did the root go? i never asked him becasue he's knida cranky...but can you help me plz?

2. Sep 28, 2005

### Tide

Your solutions certainly satisfy the original equation. Your cranky teacher might just grin if you ask him to substitute your solutions into the equation!

3. Sep 29, 2005

### Alkatran

The square root happens in in the very first step, going from ln(x^2) to 2*ln(x).

4. Sep 29, 2005

### Robokapp

oh...but it's a logaritmic property to do that. if Log(a^b) you can write b*Log(a), isn't it? i understand now that the square root only spits out the positive and that's why one root dissapears...but my real question is how can i tell that?

let's say i'm solving 10 exercises and #6 is something like this. i'm not going to stare at its graph becasue it looks so simple there's no need. I'm going to apply the properties in whatever order they come trough my mind and work my way trough algerbration. i would never have noticed this if he didn't solve it out of nowhere...so is there a way to tell that one way is better than other? because mathematically they both look correct as to applying rules...

i'm a little confused.

5. Sep 29, 2005

### VietDao29

Yup, logab = b loga
But you should note that:
$$\log x ^ 2 = \log (-x) ^ 2$$
So in this step:
Ln[(x+1)^2]=2
2Ln(x+1)=2 is wrong, after you solve this equation, you have the solution x such that x + 1 >= 0. But what if x + 1 < 0? x + 1 < 0 => (x + 1)2 > 0, so ln((x + 1)2) is still defined.
You can wrote it as:
$$\ln((x + 1) ^ 2) = 2 \Leftrightarrow \ln(|x + 1| ^ 2) = 2 \Leftrightarrow 2\ln|x + 1| = 2 \Leftrightarrow \ln|x + 1| = 1$$.
And just continue solving it, it'll give you 2 solutions.
Viet Dao,

6. Sep 29, 2005

### HallsofIvy

Your teacher may have been thinking that, since x=-e-1 is a negative number, it can't be a solution. Of course, it is not x that the ln is applied to but (x+1)2= (-e-1+1)2= (-e)2= e2 which is positive.

7. Sep 29, 2005

### Robokapp

that and the explanation above you ,with the absolute value made perfect sense. if we are to stay in the Real Axis the value plugged inside a Log or Ln has to be greater than zero.

i understand it now. i hope however you see why i had confusions...the exercise looks like a pathetic algebra 2...but it is actually let's say "controversial" becasue it takes thinknig rather than mecanically applying identities.

personally i sometimes (especially when i work in polar equations) use very very simple methods, sometimes better than what the book shows, but in other cases i complicate things beyond imagination. sometimes in geometry instead of looking for right angles so i can use pytagora i'd use Heron and the law of cosines...that's how bad i can be!