$\displaystyle 2\log_{3} y = c-\log_{3} x$

where y = 2 and x = 4

show that:

$\displaystyle y=4x^{-1/2}$

I've been able to find a value for C when y=2 and x=4 but this doesn't seem to help me at all. any help is really appreciated!

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- May 11th 2008, 02:17 PMAntProblem involving logs
$\displaystyle 2\log_{3} y = c-\log_{3} x$

where y = 2 and x = 4

show that:

$\displaystyle y=4x^{-1/2}$

I've been able to find a value for C when y=2 and x=4 but this doesn't seem to help me at all. any help is really appreciated! - May 11th 2008, 02:21 PMMoo
- May 11th 2008, 02:34 PMAnt
hi, thanks a lot for responding so fast!

$\displaystyle \log_3{xy}^2 = C$

$\displaystyle

\log_{3}xy^2 = log_3{3^c}$

at this point am I correct in saying that the log to base 3's cancel, leaving $\displaystyle xy^2 = 3^c$? - May 11th 2008, 02:41 PMMoo
- May 11th 2008, 02:56 PMAnt
okay thanks, so the rest of the problem will follow like this:

$\displaystyle xy^2 = 3^c$

$\displaystyle y = \frac{\sqrt{3^c}}{\sqrt{x}}$

$\displaystyle y = \sqrt{3^c} * x^{-1/2}$

Where C can be found:

$\displaystyle xy^2=3^c$

$\displaystyle 16=3^c$

$\displaystyle lg{16}=C * lg{3}$

$\displaystyle \frac{lg{16}}{lg{3}}=c$