$\displaystyle 2\log_{3} y = c-\log_{3} x$
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!
$\displaystyle 2\log_{3} y = c-\log_{3} x$
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!
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:
$\displaystyle xy^2=3^c$
$\displaystyle 16=3^c$
$\displaystyle lg{16}=C * lg{3}$
$\displaystyle \frac{lg{16}}{lg{3}}=c$