# Logmarithmic property woes and algebraic confusion

1. Dec 23, 2008

### kenewbie

I'm having trouble coming to terms with the following:

ln x^n = n ln x

Which is all nice and well until I tried

ln(x + 6) = 2 ln x

which is true for x = 3.

however, 2 ln x = ln x^2 .. so

ln(x + 6) = ln x^2

.. which is true for x = 3 AND x = -2.

so the two different ways to write the ln expression are not really equivalent.

I'm not quite sure what to make of this. It just feels wrong.

k

2. Dec 23, 2008

### Vagrant

Why is ln(x+6)=2 ln x ? Is this an equation you have to solve to get the values of x?

3. Dec 23, 2008

### kenewbie

It is an example of an equation in which have different solutions depending on which form of the logarithm you use. It has one solution when the right side is 2 ln x, and two solutions when the right side is ln x^2. Yet 2 ln x is supposed to be equivalent to ln x^2

k

4. Dec 23, 2008

### Vagrant

The natural logarithm ln(x) is defined for x > 0. So when you use the property ln x^n = n ln x, you loose the negative solution of x.

5. Dec 23, 2008

### kenewbie

.. so what you are saying is that ln x^n != n ln x ?

if you loose solutions, I don't see how one could call them equivalent.

k

6. Dec 23, 2008

### snipez90

No, the set of solutions to an equation contain solutions for which the equation (the one you are trying to solve) holds. In this case, the equation is ln(x + 6) = 2 ln x.

Whenever you square a quantity, there is a certain loss of information. This is not a big deal if you are careful. Once you solve the quadratic and get x = 3 and x = -2, you should check that both are indeed solutions. It's clear that ln(-2) isn't even defined, so x = -2 is an extraneous solution, which has to thrown out since the original equation does not hold.

The moral of the story is to always check to make sure your solutions do indeed work, especially when squares are involved. Of course, it's also help to remember what the domain of various functions is.

7. Dec 23, 2008

### snipez90

They are equivalent provided that the argument is positive and real, i.e., x > 0.

8. Dec 23, 2008

### uart

Yes kenewbie, any time you do an operation that is only applicable for certain values of "x" you can potentially lose solutions. Lets take a really trivial example to demonstrate, say you wanted to solve x^2 - x = 0.

Divide by x to get x - 1 = 0 so x=1 is the solution. However we know there are really two solutions, x=1 and x=0, to the original equation. So what happened to the other solution? I'm sure you weren't fooled by this one right, when I divided by "x" I should have explicitly stated for x not equal to zero, so that's how the other solution got lost.

Last edited: Dec 23, 2008
9. Dec 23, 2008

n ln x = ln xn provided that both sides are defined (i.e. x > 0). That means, you can replace n ln x with ln xn wherever you see it, but to go the other way, you have to make sure x > 0.

For example, 2 ln x only makes sense for x > 0, whereas ln x2 makes sense for any nonzero x.

10. Dec 23, 2008

### kenewbie

uart: that makes a lot of sense, thanks.

k