Question 187879
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\, (x) = \log_3\,\left(\frac{1}{x}\right)+4]


9 certainly could be (and is) the value of <i>x</i> that makes this equation a true statement, but -9 cannot be.  The domain of the <i>log</i> function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \{x\,|\,x\,\in\,\R,\ x > 0}]


So if you were to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\, (-9) = \log_3\,\left(\frac{1}{-9}\right)+4]


What you would really be saying is: "Some undefined thing is equal to some other undefined thing plus 4."  A patently absurd thing to say by any measure.


Having said all of that, here's how to do the problem correctly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\, (x) = \log_3\,\left(\frac{1}{x}\right)+4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\, (x) - \log_3\,\left(\frac{1}{x}\right)=4]


Now, use "The log of the quotient is the difference of the logs"


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\, (x) - ( \log_3\,(1) - \log_3\,(x)) =4]


Remove the parentheses and use *[tex \Large \log_b\,(1) = 0] to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\, (x) + \log_3\,(x) =4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2 \log_3\,(x) =4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\,(x) =2]


Since: *[tex \Large \log_b\,(x) = y \ \ \Rightarrow\ \ b^y = x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\,(x) =2\ \ \Rightarrow\ \ 3^2 = x \ \ \Rightarrow\ \ x = 9]


Where whomever gave you the answer of -9 may have gone wrong is they might have said:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2 \log_3\,(x) =4\ \ \Rightarrow\ \ \log_3\,(x^2)=4\ \ \Rightarrow\ \ 3^4 = x^2 \ \ \Rightarrow\ \ x^2 = 81]


And the roots of *[tex \Large x^2 = 81] are 9 and -9, but you still have to go back to the original equation and consider any domain restrictions.  Therefore, you must exclude the negative root.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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