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I am a little confused as to why sometimes they will treat $x$ in $\ln x$ as $\ln ab$ and other times its treated as $\ln x$.... To be more clear, here is an example:

$2\ln(3x^2-1)$ and $\ln2x$. Now, in $2\ln(3x^2-1)$ the $3x^2-1$ becomes $x$ and is written as $2\cdot1/x\cdot x'$. My question is why? because this could have been treated as $\ln3x^2/\ln1$.

On the other hand, $\ln2x$ they solve as $\ln2+\ln x$ then differentiate it.... So my question is: When do they choose to apply the log rules and when not to? Will not applying the rule give you the wrong answer? In the case of $2\ln(3x^2-1)$ could they have done it differently as oppose to treating $3x^2-1$ as $x$.

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I think you may want to rewrite parts of this. It's incomprehensible as it is now. I assume lnx is $\ln x$, but what's lnab? And the second paragraph is completely unreadable. –  Mike Dec 8 '11 at 15:48

3 Answers 3

up vote 1 down vote accepted

I hope that the following comments settle some of your questions. Unfortunately, it is not clear what notation you use for the Chain Rule.

1) Suppose that $H(x)=\ln(5x)$. We want to find $H'(x)$. Note that $\ln(5x)$ is "function of a function of $x$." Specifically, $\ln(5x)=f(g(x)$, where $g(x)=5x$ and $f(u)=\ln(u)$.

Therefore, by the Chain Rule, we have $H'(x)=g'(x)f'(g(x))$. Note that $g'(x)=5$ and $f'(u)=\dfrac{1}{u}$. So $f'(g(x))=\dfrac{1}{5x}$.

Now put things together. We have $H'(x)=(5)\dfrac{1}{5x}$. This simplifies to $\dfrac{1}{x}$.

In this case, we could have avoided the Chain Rule, by noting that $H(x)=\ln(5x)=\ln(5)+\ln(x)$. Differentiate, using the fact that $\ln(5)$ is a constant and therefore its derivative is $0$. We get $H'(x)=\dfrac{1}{x}$.

2) Let $H(x)=\ln(3x^2-1)$. We want $H'(x)$. Note that $H(x)=f(g(x))$ where $g(x)=3x^2-1$ and $f(u)=\ln(u)$. Use the Chain Rule like before. We have $g'(x)=6x$ and $f'(u)=\dfrac{1}{u}$. Putting things together like before, we get $H'(x)=\dfrac{6x}{3x^2-1}$. No useful simplification is available.

Important note: It looks as if your post asks why we can't treat $\ln(3x^2-1)$ as $\dfrac{\ln(3x^2)}{\ln(1)}$.

The reason that we can't is that $\ln(3x^2-1)$ is not equal to $\dfrac{\ln(3x^2)}{\ln(1)}$. The logarithm of a product is a sum of logarithms. the logarithm of a quotient is a difference of logarithms. So for example, $\ln((x^2+1)(x^2+5))=\ln(x^2+1)+\ln(x^2+5)$, and $\ln(7/5)=\ln(7)-\ln(5)$.

But the logarithm of a sum can't be "simplified" in a pleasant way, and neither can the logarithm of a difference. You can check for yourself, with a calculator, that $\ln(7-2)$ is not equal to $\dfrac{\ln(7)}{\ln(2)}.$

If you use the FALSE "rules" $\ln(a+b)=\ln(a)+\ln(b)$ or $\ln(a-b)=\dfrac{\ln(a)}{\ln(b)}$, it will cause you grief over and over.

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$\ln(3x^2-1)$ is certainly not the same as $\ln(3x^2)/(\ln 1)$.

Correct: $\ln\dfrac ab=(\ln a) - (\ln b)$

Incorrect: $\ln(a-b) = \dfrac{\ln a}{\ln b}$

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thank you all! i would logically think that the sum or difference rules for ln exist but they dont! –  Raynos Dec 8 '11 at 20:54

Now, in $2\ln(3x^2-1)$ the $3x^2-1$ becomes $x$ and is written as $2\cdot1/x\cdot x'$.

I'll assume you mean it's treated as another variable (not $x$), perhaps $u$, and you differentiate $2\ln u$ to obtain $2\cdot\frac1uu'$ (which is then $2\cdot\frac1{3x^2-1}6x$.

The answer is that it doesn't matter. You can do either method in either case. But whereas $\ln 2x$ is easy to split up into $\ln2+\ln x$, $\ln(3x^2-1)$ is not as easy to split up — see Michael Hardy's answer — so they don't split it up. (But if you'd find a way to split it up, you could.)

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