SOLUTION: w=cr^(-2) solve for r

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: w=cr^(-2) solve for r      Log On


   



Question 403368: w=cr^(-2) solve for r
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
w=c%2Ar%5E%28-2%29
"Solve for r" means we want r by itself on one side of the equation. So we want to "remove" the c in front and the exponent of -2. First we will divide by c which, as you will see, will "move" the c over to the other side of the equation:
%28w%29%2Fc=%28c%2Ar%5E%28-2%29%29%2Fc
w%2Fc=%28cross%28c%29%2Ar%5E%28-2%29%29%2Fcross%28c%29
w%2Fc=+r%5E%28-2%29
Now we have an equation that tells us what r%5E%28-2%29 is. We want an equation that tells us what "r" is. The exponent on "r" is a 1. (Any "invisible" exponent is always a 1.) So we want to find a way to change the exponent from a -2 into a 1. To do this we will combine several ideas:
  • It is OK to raise both sides of an equation to the same power.
  • The rule for exponents when raising a power to a power is to multiply the exponents.
  • Multiplying reciprocals always results in a 1!

So if we raise both sides of the equation to the reciprocal of -2 power, the exponent on r will turn into a 1! The reciprocal of -2 is -1/2. So we will raise both sides of the equation to the -1/2 power:
%28w%2Fc%29%5E%28-1%2F2%29=+%28r%5E%28-2%29%29%5E%28-1%2F2%29
On the right side we get r%5E1 or just r, exactly as we planned:
%28w%2Fc%29%5E%28-1%2F2%29=+r
This may be an acceptable form for the answer. Or we could do some things on the left side:
%28w%2Fc%29%5E%28%28-1%29%2A%281%2F2%29%29=+r
%28%28w%2Fc%29%5E%28-1%29%29%5E%281%2F2%29=+r
%28c%2Fw%29%5E%281%2F2%29=+r
sqrt%28c%2Fw%29=+r
sqrt%28%28c%2Fw%29%28w%2Fw%29%29=+r
sqrt%28cw%2Fw%5E2%29=+r
sqrt%28cw%29%2Fsqrt%28w%5E2%29=+r
sqrt%28cw%29%2Fw=+r
This may be the preferred form for the answer.