anonymous
  • anonymous
Determine whether the function has an inverse function. : y=lnx^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\ln(x^2)~~\text{or}~~(\ln x)^2~~?\]
anonymous
  • anonymous
It doesnt have any perenthesis
anonymous
  • anonymous
If that's the case, I'd assume the first one.

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More answers

anonymous
  • anonymous
|dw:1377745104389:dw|
anonymous
  • anonymous
The usual procedure is to flip the variables, then solve for \(y\). \[y=\ln(x^2)~~\Rightarrow~~x=\ln(y^2)\]
anonymous
  • anonymous
got that
anonymous
  • anonymous
but how fo you solve for y?
anonymous
  • anonymous
\[x=\ln(y^2)\\ x=2\ln y\\ \frac{x}{2}=\ln y\\ y=\cdots \]
anonymous
  • anonymous
In my book, the answer says there is no inverse... Im assuming that's because if you graph the function it doesn't pass the horizontal/one-to-one test But how would I know that by solving it algebraically like that I have no idea how to graph an equation like that either
zepdrix
  • zepdrix
https://www.desmos.com/calculator/fw7vmht6fc Oh I see what you mean. Look at the graph I provided. Our red function, when reflected over y=x (in green) gives us the inverse function in purple. See how we lost one of the branches? :O (Otherwise it wouldn't pass the horizontal line test, as you said.) Algebraically? Hmmm..
zepdrix
  • zepdrix
I guess the problem is ~ when we apply our rule of logs:\[\Large y=\ln(x^2) \qquad\to\qquad y=2\ln x\]We lose information. The domain of our x changes so we can't include negative numbers anymore.
zepdrix
  • zepdrix
I think we would want to write it like this to preserve those values.\[\Large y=2\ln|x|\]Then we can perform the inverse stuff from here.
zepdrix
  • zepdrix
\[\Large x=2\ln|y|\]And then do some stuff...\[\Large \frac{1}{2}x=\ln|y|\]Exponentiating both sides,\[\Large |y|=e^{1/2x}\]
zepdrix
  • zepdrix
\[\Large y=\pm e^{1/2x}\]
zepdrix
  • zepdrix
I `think` that's how you're supposed to approach that... with the absolute bars. It seems to get us in the right direction at least.
zepdrix
  • zepdrix
And from there we would determine that there is no inverse, due to that plus/minus sign.
anonymous
  • anonymous
Thank you!

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