SOLUTION: Solve. include approximations to the nearest thousandth.
log(x-3) + logx = 1
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Question 235044: Solve. include approximations to the nearest thousandth.
log(x-3) + logx = 1
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ +\ \log(x)\ =\ 1)
The sum of the logs is the log of the product, that is:
 + \log_b(y) = \log_b(xy))
Therefore
\ =\ 1)
If the base is unspecified, base 10 is understood, so:
\ =\ 1)
Apply the definition of logarithms
 \ \ \Rightarrow\ \ b^y = x)
\ \Rightarrow\ \ 10^1\ =\ x^2\ -\ 3x)
Put that into standard form and you have a factorable quadratic equation:
(x\ +\ 2)\ =\ 0)
or 
But 
is not in the domain of )
, therefore you must exclude this root.
Check
\ =\ \log(2)\ \approx\ 0.3010)
\ \approx\ 0.6990)
John
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