# Problem involving logs

• May 11th 2008, 02:17 PM
Ant
Problem 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 PM
Moo
Hello,

Quote:

Originally Posted by Ant
$\displaystyle 2\log_{3} y = c-\log_{3} x$

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!

$\displaystyle a \log b=\log b^a$

$\displaystyle \log a+\log b=\log ab$

--> $\displaystyle 2\log_{3} y=\log_3 y^2$

----> $\displaystyle \log_3 y^2+\log_3 x=c$

------> $\displaystyle \log_3 xy^2=c$

--------> $\displaystyle xy^2=3^c$

etc :)
• May 11th 2008, 02:34 PM
Ant
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 PM
Moo
Quote:

Originally Posted by Ant
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 $\displaystyle xy^2 = 3^c$?

Yes ! (Clapping)
• May 11th 2008, 02:56 PM
Ant
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:

$\displaystyle xy^2=3^c$

$\displaystyle 16=3^c$

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

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