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In the following equation how come $\ln(x)$ is part of the factor? I would understand if $\ln(x^2)$. But how can you include $\ln(x)$ into the factoring? If you expanded this would $\ln(x)$ be unaffected from $x^2$? $$x^2 + 3x^2 \ln(x) = x^2 ( 1 + 3\ln(x))$$

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3 Answers 3

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Consider $\ln(x)$ as its own object. When you pull a factor of $x^2$ out of each term, the first term leaves a $1$ and the second term a $3\ln(x)$. Alternatively, going in the other direction, when you distribute the $x^2$ to both terms the second is $x^2\times2\ln(x)$ which has nothing to do with $\ln(x^2)$.

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  • $\begingroup$ So $ln(x)$ is simply uneffected by multiplication? How would you even get $ln(x^2)$ $\endgroup$ Feb 21, 2014 at 1:54
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    $\begingroup$ You could take the logarithm of $x^2$. There are also various properties of logarithms. There is the fact that $\ln(x^a)=a\ln(x)$ so getting something that looks like $\ln(x^2)$ isn't really that weird, but is off by a factor of two of what you have here. $\endgroup$
    – jazzwhiz
    Feb 21, 2014 at 2:03
  • $\begingroup$ @user3268003 If you want to get $ln(x^2)$, you need to have $2ln(x)$ since $log_a(b^c) = c × log_a(b)$. $\endgroup$ Feb 21, 2014 at 2:03
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Have you ever factored an expression using a substitution (e.g. $2a^6-a^3-6$). You can factor $x^2+3x^2\ln(x)$ using a substitution as well. Let $\ln(x)=a$. Then $x^2+3x^2\ln(x)=x^2+3x^2a$. Factoring gives us $x^2(1+3a)$. Substitute $\ln(x)$ back into $a$. The answer is: $$\boxed{x^2+3x^2\ln(x)=x^2\left[1+3\ln(x)\right]}$$

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  • $\begingroup$ This is the way I would have explained it. $\endgroup$
    – Lubin
    Feb 23, 2014 at 4:12
  • $\begingroup$ So basically, until I am in a situation to to use properties of logarithms, I should just treat logarithms as a dummy variable for simplicity. That makes a lot of sense; simple and useful. Thanks mate $\endgroup$ Feb 23, 2014 at 12:19
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I'm really not sure where you're getting $\ln (x^2)$ from, but here's an attempt to clear this up:

Suppose that we just said $Y = \ln(x)$. Then we could write this as $$ x^2 + 3x^2 \ln(x) = x^2 + 3x^2 Y $$ Now, $Y$ is just a number. All we need to do is treat it as such, and factor the expression. So, we have $$ x^2 + 3x^2 Y = x^2 + x^2\cdot (3Y) = x^2(1 + 3Y) = x^2(1 + 3\ln(x)) $$

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