Solve for "r"

[mjd7132]

I'm stumped.

The task is to solve for r.
A, B, and n are constants, but I'm not given what their value is.


0=(A/r[sup:3fqok0k5]2[/sup:3fqok0k5])-(Bnr[sup:3fqok0k5]n-1[/sup:3fqok0k5]/r[sup:3fqok0k5]2n[/sup:3fqok0k5])


Heeeeeelp please!

 

[mmm4444bot]

Thanks for clarifying.

MY EDIT: Deleted comments.

 

[mjd7132]

My apologies, I adjusted the problem from E=f(x) to 0=f(x), which is the appropriate form.

I cannot figure out how to pull "r" out of the equation to solve for it when "r"'s exponents are so complex!



Please point me in the right direction if anyone has any insight.

 

[mmm4444bot]

Start by getting r out of the denominators of those two fractions. Multiply both sides by r^2.

That gives:

A = B * n * r^(1 - n)

Can you go from here?

 

[mmm4444bot]

BTW: You'll need to be at ease with the following properties of exponents and logarithms.

b^n * b^m = b^(n + m)

b^n/b^m = b^(n - m)

ln(r^c) = c*ln(r)

e^ln(x) = x

If you still need help with anything, please show whatever work you can or explain why you're stuck or confused.

I don't really know what kind of help that you need because I don't know what you already know about this stuff.

 

[mmm4444bot]

Did you go to sleep?

 

[mmm4444bot]

\(\displaystyle \frac{A}{r^2} - \frac{B \cdot n \cdot r^{n - 1}}{r^{2n}} \;=\; 0\)

\(\displaystyle r^2 \cdot \left( \frac{A}{r^2} - \frac{B \cdot n \cdot r^{n - 1}}{r^{2n}} \right) \;=\; r^2 \cdot 0\)

\(\displaystyle \frac{r^2}{1} \cdot \frac{A}{r^2} - \frac{r^2}{1} \cdot \frac{B \cdot n \cdot r^{n - 1}}{r^{2n}} \;=\; 0\)

\(\displaystyle A - \frac{B \cdot n \cdot r^{n - 1} \cdot r^2}{r^{2n}} \;=\; 0\)

Can you use the property above to multiply the two powers of r in the numerator together?

Then use the property above for a ratio of powers, to get the single power of r^(1 - n) shown in my post above.

Gosh, did I just join the "Elite" members' club ? (Finally, I'm just as good as Denis.)

 

[mjd7132]

Thank you very much for getting me started! I have to go to class, but I'll continue to work on the problem there. You're awesome!

 

[mmm4444bot]

You're awesome!

I know.

 

[Denis]

Re:

ln(r^c) = c*ln(r)
e^ln(x) = x

Don't think those are required; once you're at: A = B * n * r^(1 - n), then:

r^(1 - n) = A / (Bn)

r = [A / (Bn)]^[1 / (1 - n)]

Mais oui?

 

[mmm4444bot]

ln(r^c) = c*ln(r)
e^ln(x) = x

Don't think those are required; once you're at: A = B * n * r^(1 - n), then:

r^(1 - n) = A / (Bn)

r = [A / (Bn)]^[1 / (1 - n)]

Mais oui?

Ah, yes. Raising both sides to the power of 1/(1 - n) also works.