Problem involving logs

**Ant** · May 11th 2008, 02:17 PM

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

where y = 2 and x = 4

show that:

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

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!

**Moo** · May 11th 2008, 02:21 PM

Hello,

Originally Posted by Ant

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

where y = 2 and x = 4

show that:

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

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!

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

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

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

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

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

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

etc

**Ant** · May 11th 2008, 02:34 PM

hi, thanks a lot for responding so fast!

$\log_3{xy}^2 = C$
$<br /> \log_{3}xy^2 = log_3{3^c}$

at this point am I correct in saying that the log to base 3's cancel, leaving $xy^2 = 3^c$ ?

**Moo** · May 11th 2008, 02:41 PM

Originally Posted by Ant

hi, thanks a lot for responding so fast!

$\log_3{xy}^2 = C$
$<br /> \log_{3}xy^2 = log_3{3^c}$

at this point am I correct in saying that the log to base 3's cancel, leaving $xy^2 = 3^c$ ?

Yes !

**Ant** · May 11th 2008, 02:56 PM

okay thanks, so the rest of the problem will follow like this:

$xy^2 = 3^c$

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

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

Where C can be found:

$xy^2=3^c$

$16=3^c$

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

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

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