If f(x) = log(x^2) and g(x) = (x/10), then how can we express f(g(x))?

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- Apr 19th 2009, 07:09 AMmagentaritaExpress f(g(x))
If f(x) = log(x^2) and g(x) = (x/10), then how can we express f(g(x))?

- Apr 19th 2009, 07:53 AMrunning-gag
Hi

$\displaystyle f(x) = \log(x^2) = 2 \log |x|$

$\displaystyle f(g(x)) = f\left(\frac{x}{10}\right) = 2 \log \left(\frac{|x|}{10}\right) = 2 (\log |x| - \log 10) = 2(\log |x| - 1)$

or $\displaystyle f(g(x)) = f\left(\frac{x}{10}\right) = \log \left(\frac{x^2}{100}\right) = \log (x^2) - \log 100 = \log (x^2) - 2$ - Apr 20th 2009, 04:41 AMmagentaritawhere???
- Apr 20th 2009, 05:57 AMstapel
A

**log rule**lets you take powers out front as multipliers. This is fine for stuff like:

. . . . .$\displaystyle \log(3^4)\, =\, 4\, \log(3)$

In the above, you know that 3 is positive, so it's okay to take the 4 out front as a multiplier. However, in the listed case, a power is being taken off a variable. We don't know whether or not this variable is positive, negative, or zero. (Well, we'll assume it's not zero, since the square of zero, being still zero, is not allowed inside the log.)

Since logs are defined only for positive inputs, then we can't have a negative value for the argument. When the x was squared, this was fine: even if x is negative, the square of x is positive.

However, once the square was taken out front as a "times two", we lost our assurance regarding the sign of x. To keep things "kosher", we have to use absolute-value bars, thus ensuring that the argument of the log remains positive.

(Wink) - Apr 20th 2009, 09:03 PMmagentaritaok....