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Error in One of the Laws of Logarithms?

Date: 05/02/2002 at 13:59:30
From: Charles Pence
Subject: Error in one of the laws of logarithms?

We were discussing a problem in precalculus today and seemed to 
discover a basic flaw in one of the exponent laws.  Recall:

   log(x^2) = 2log(x)

It is also a fact that the log functions have a domain restriction on 
all values less than or equal to zero.

However:

For log(x^2), the only value that is restricted (less than or equal 
to zero) is zero itself (log(0^2) = log(0)).

For 2log(x), which should supposedly be the same function by the law 
stated above, there are restrictions on all values of X less than or 
equal to zero.

Basically, what it comes down to is that negative numbers are an 
acceptable input for the function before you apply the law, and 
negative numbers are no longer an acceptable input after you apply 
the law. This means that the two functions are not the same, and 
inherently disproves that law of logarithims. Doesn't it?

We're thoroughly perplexed, and resorted to the Internet for 
assistance.

Thanks,
Charles Pence


Date: 05/02/2002 at 16:12:49
From: Doctor Peterson
Subject: Re: Error in one of the laws of logarithms?

Hi, Charles.

You haven't shown that the law is wrong, but only that it has an 
implicit restriction:

    log(a^b) = b log(a)
    for all a and b for which both logarithms are defined

If a is negative and b even, then the left side is defined but the 
right side is not. You are taking the property to mean that the 
function on the left is identical to the function on the right, 
including having the same domain; but that's not what it means. It is 
only a pointwise identity (true for one pair of values at a time), not 
a statement about the two functions as a whole.

The same can be said of the other logarithm identities, such as

    log(ab) = log(a) + log(b)

where, if a and b are both negative, the left side is defined but the 
right is not.

We even face the same problem with simpler facts:

    (sqrt(x))^2 = x

is true whenever sqrt(x) is defined; it is not wrong just because 
there are values of x for which only the right side is defined. We 
just have to clearly state the restriction: "for all x >= 0" .

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Logs

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