Logmarithmic property woes and algebraic confusion

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Discussion Overview

The discussion revolves around the properties of logarithms, specifically the logarithmic identity ln(x^n) = n ln(x) and its implications when applied to different forms of logarithmic equations. Participants explore the conditions under which these identities hold true and the potential loss of solutions when manipulating logarithmic expressions.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion over the identity ln(x^n) = n ln(x) when applied to the equation ln(x + 6) = 2 ln(x), noting that it appears to yield different solutions.
  • Another participant questions whether ln(x + 6) = 2 ln(x) is an equation that needs solving for x.
  • It is noted that the equation has one solution when using 2 ln(x) and two solutions when using ln(x^2), raising concerns about the equivalence of these forms.
  • A participant points out that the natural logarithm is defined only for x > 0, suggesting that this restriction leads to the loss of solutions.
  • There is a discussion about the implications of squaring a quantity, which can lead to extraneous solutions, emphasizing the importance of checking solutions against the original equation.
  • Another participant clarifies that ln(x^n) = n ln(x) is valid only when x is positive and real, highlighting the need for caution in applying logarithmic identities.
  • A trivial example is provided to illustrate how dividing by a variable can lead to lost solutions, reinforcing the idea that operations must be carefully considered.
  • It is stated that n ln(x) = ln(x^n) holds true under the condition that both sides are defined, specifically for x > 0.

Areas of Agreement / Disagreement

Participants generally agree that the logarithmic identities can lead to different solutions depending on the context and the values of x. However, there is no consensus on the implications of these differences, and the discussion remains unresolved regarding the equivalence of the logarithmic forms in all cases.

Contextual Notes

Limitations include the dependence on the domain of the logarithmic function, as well as the potential for extraneous solutions when manipulating equations involving logarithms.

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

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.
 
kenewbie said:
.. 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

They are equivalent provided that the argument is positive and real, i.e., x > 0.
 
Yes kenewbie, any time you do an operation that is only applicable for certain values of "x" you can potentially lose solutions. Let's 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:
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
uart: that makes a lot of sense, thanks.

k
 

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