# Logarithm Help

• Sep 13th 2009, 03:52 AM
user_5
Logarithm Help
If $\log_{10} (x - 2) - 3 \log_{10} (2x) = 1 - \log_{10} (y)$, then y is equal to
• Sep 13th 2009, 04:01 AM
mr fantastic
Quote:

Originally Posted by user_5
If $\log_{10} (x - 2) - 3 \log_{10} (2x) = 1 - \log_{10} (y)$, then y is equal to

Using basic log rules:

The left hand hand side can be written $\log_{10} \frac{x - 2}{8x^3}$.

The right hand side can be written $\log_{10} 10 - \log_{10} y = \log_{10} \frac{10}{y}$.

Therefore ....
• Sep 13th 2009, 04:08 AM
user_5
Oh, thanks I forgot that 1 can be rewritten like that.
It makes sense now.

Is the answer is $y= \frac {80x^3}{x-2}$ ?
• Sep 13th 2009, 04:56 AM
Finley
Yes, well done :)

But don't forget the restrictions on X, for the equation to produce a positive number (necessary for a log).

$x>2$ and $x<0$