If $\displaystyle \log_{10} (x - 2) - 3 \log_{10} (2x) = 1 - \log_{10} (y)$, then y is equal to
Last edited by mr fantastic; Sep 13th 2009 at 03:59 AM. Reason: Fixed latex
Follow Math Help Forum on Facebook and Google+
Originally Posted by user_5 If $\displaystyle \log_{10} (x - 2) - 3 \log_{10} (2x) = 1 - \log_{10} (y)$, then y is equal to Using basic log rules: The left hand hand side can be written $\displaystyle \log_{10} \frac{x - 2}{8x^3}$. The right hand side can be written $\displaystyle \log_{10} 10 - \log_{10} y = \log_{10} \frac{10}{y}$. Therefore ....
Oh, thanks I forgot that 1 can be rewritten like that. It makes sense now. Is the answer is $\displaystyle y= \frac {80x^3}{x-2} $ ?
Yes, well done But don't forget the restrictions on X, for the equation to produce a positive number (necessary for a log). $\displaystyle x>2$ and $\displaystyle x<0$
Last edited by Finley; Sep 13th 2009 at 05:11 AM.
View Tag Cloud