Question

I need help expressing y as a function of x.
2ln y=-1/2ln x+1/3ln(x^2+1)+ln c

Answer 1

Hey Squirrel :) Welcome to OpenStudy!

Answer 2

y a function of x? Hmmm... let's see...

Answer 3

\[\Large\rm 2\ln y=-\frac{1}{2}\ln x+\frac{1}{3}\ln(x^2+1)+\ln c\]Let's divide both sides by 2 to get isolate our `ln y`,\[\Large\rm \ln y=-\frac{1}{4}\ln x+\frac{1}{6}\ln(x^2+1)+\frac{1}{2}\ln c\]

Answer 4

We would like to condense this down to one or two logs on the right side, then we can do some exponentiation.

Answer 5

So we need to apply some log rules.

Answer 6

\[\Large\rm \ln y=\ln\left(\frac{1}{\sqrt[4]{x}}\right)+\frac{1}{6}\ln(x^2+1)+\frac{1}{2}\ln c\]Do you understand what I did in that first step?
Or should I slow it down a lil bit?

Answer 7

what log rules did you apply?

Answer 8

\[\Large\rm a \log(b)=\log(b^a)\]Since the thing out front was a fraction, I rewrote it as a root.\[\Large\rm -\frac{1}{4}\ln(x)=\ln\left(x^{-1/4}\right)=\ln\left(\frac{1}{x^{1/4}}\right)=\ln\left(\frac{1}{\sqrt[4]{x}}\right)\]

Answer 9

oh ok thank you now that step is clear

Answer 10

do we ever cancel the ln?

Answer 11

There is a way we can deal with the logs, but not yet.
We first to get it into this form:

ln(y) = ln(stuff)

We would like a single log on the right side.

Answer 12

After you apply that same log rule to the other two terms,
you get something like this:\[\Large\rm \ln y=\ln\left(\frac{1}{\sqrt[4]{x}}\right)+\ln\left(\sqrt[6]{x^2+1}\right)+\ln\sqrt{c}\]

Answer 13

Then we need to apply this rule:\[\Large\rm \log(a)+\log(b)=\log(ab)\]

Answer 14

is that the product one?

Answer 15

yes

Answer 16

and we combine it?

Answer 17

Yes, so combining the first two terms should give us something like this,\[\Large\rm \ln y=\ln\left(\frac{\sqrt[6]{x^2+1}}{\sqrt[4]{x}}\right)+ln\sqrt{c}\]Yah?

Answer 18

ok yes

Answer 19

we cannot reduce the 6 and 4?

Answer 20

No. The numerator and denominator do not have the same base.

Answer 21

ok is there anything i can do here in terms of reducing?

Answer 22

So apply that log rule one more time,
\[\Large\rm \ln y=\ln\left(\frac{\sqrt[6]{x^2+1}~\sqrt c}{\sqrt[4]{x}}\right)\]

Answer 23

And now it's in the form that we wanted:\[\Large\rm \ln y=\ln(stuff)\]

Answer 24

We can exponentiate each side,
Or maybe easier yet, just recognize that each side is inside of a log,
and each log is equal to one another,
then the insides of those logs have to be equal as well.\[\Large\rm y=\frac{\sqrt[6]{x^2+1}~\sqrt{c}}{\sqrt[4]{x}}\]

Answer 25

so that gets rid of the ln?

Answer 26

Yes.
If you prefer to be a little bit more careful about it,
you can exponentiate each side:\[\Large\rm \ln y=\ln(stuff)\]\[\Large\rm e^{\ln y}=e^{\ln(stuff)}\]Since the exponential and log are inverse operations of one another they essentially "undo" one another,\[\Large\rm y=stuff\]

Answer 27

ok makes sense, btw thank you thank you this has been so helpful

Answer 28

is that the final answer?