SOLUTION: log base 3 (x) = log base 3 (1/x)+4
i understand that log base 3 (1/x) is subtracted so the problem could look something like this log base 3 (x) -log base 3 (1/x)=4
have one
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Question 187879: log base 3 (x) = log base 3 (1/x)+4
i understand that log base 3 (1/x) is subtracted so the problem could look something like this log base 3 (x) -log base 3 (1/x)=4
have one question doesnt 1 and x become log base 3 (1)- log base 3 (x)
the answer is 9, -9 but i dont know how to set up the problem i am stuck
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
 = \log_3\,\left(\frac{1}{x}\right)+4)
9 certainly could be (and is) the value of x that makes this equation a true statement, but -9 cannot be. The domain of the log function is:
So if you were to say:
 = \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:
 - \log_3\,\left(\frac{1}{x}\right)=4)
Now, use "The log of the quotient is the difference of the logs"
 - ( \log_3\,(1) - \log_3\,(x)) =4)
Remove the parentheses and use  = 0)
to write:
 + \log_3\,(x) =4)
 =4)
 =2)
Since:  = y \ \ \Rightarrow\ \ b^y = 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:
 =4\ \ \Rightarrow\ \ \log_3\,(x^2)=4\ \ \Rightarrow\ \ 3^4 = x^2 \ \ \Rightarrow\ \ x^2 = 81)
And the roots of 
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
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