# Product rule for logarithms

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• August 14th 2010, 12:20 PM
zg12
Product rule for logarithms
Have a silly question.

In the proof that ln(ax)=ln(a) + ln(x)

we show that ln(ax) and ln(x) have the same derivative, therefore they differ by a constant; ln(a)=ln(x) + C

The book then sets x=1, and says C=ln(a), and the rule is proved. But I get C=ln(a)-ln(x), and that x=1 is a special case.

What's my mistake?
• August 14th 2010, 12:39 PM
Plato
Look at it this way.
$\ln(x)=\ln(a)-C$ so $\ln(1)=0=\ln(a)-C$.
Therefore $C=\ln(a).$
• August 14th 2010, 12:56 PM
zg12
What I'm not understanding is how we can set x= any number, without loosing the generality of the rule. I see that ln(a)=ln(x) + C is true for any value of x, but if we choose x=2 instead of x=1 then we get C=ln(a)-ln(2) instead of C=ln(a)
• August 14th 2010, 01:02 PM
zg12
Is it because that C + ln (2) is just another constant?

C+ln(2)=ln(a), which equals C=ln(a) ????????
• August 14th 2010, 01:31 PM
Plato
No that is not the point. (BTW: I don't like that proof).
But the fact is $\ln(1)=0$. Therefore, it must be the case that $C=\ln(a)$.
• August 14th 2010, 01:42 PM
Archie Meade
Maybe the diagram will help show why x=1 is chosen (for the case of a=2).

Remember, in order to discover the constant C, you must eliminate x.
Setting x=1 causes ln(x) to be zero, eliminating x.

Also, you have a typo in your post

ln(a)=ln(x)+C should read ln(ax)=ln(x)+C.

Hence C=ln(ax)-ln(x) but only "removing" x shows the value of C.

Setting x to 1,

$ln(a)-ln(1)=C$

$ln(a)-0=C$
• August 15th 2010, 05:33 AM
zg12
Resolved. I woke up this morning and can't even remember what my confusion was over.