mathdad Full Member Joined Apr 24, 2015 Messages 941 May 14, 2019 #1 Given R = (CL)/r^4, solve for r. Solution: Given R = (CL)/r^4. Multiplying both sides by r^4, I got: Rr^4 = CL r^4 = CL/R sqrt{r^4} = sqrt{CL/R} r^2 = sqrt{CL/R} sqrt{r^2} = sqrt{sqrt{CL/R}} r = sqrt{sqrt{CL/R}} Yes?
Given R = (CL)/r^4, solve for r. Solution: Given R = (CL)/r^4. Multiplying both sides by r^4, I got: Rr^4 = CL r^4 = CL/R sqrt{r^4} = sqrt{CL/R} r^2 = sqrt{CL/R} sqrt{r^2} = sqrt{sqrt{CL/R}} r = sqrt{sqrt{CL/R}} Yes?
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 May 14, 2019 #2 When you get to: [MATH]r^4=\frac{CL}{R}[/MATH] I would next write: [MATH]r=\pm\sqrt[4]{\frac{CL}{R}}[/MATH] or: [MATH]r=\pm\left(\frac{CL}{R}\right)^{\frac{1}{4}}[/MATH] This is assuming we are only after real solutions.
When you get to: [MATH]r^4=\frac{CL}{R}[/MATH] I would next write: [MATH]r=\pm\sqrt[4]{\frac{CL}{R}}[/MATH] or: [MATH]r=\pm\left(\frac{CL}{R}\right)^{\frac{1}{4}}[/MATH] This is assuming we are only after real solutions.
mathdad Full Member Joined Apr 24, 2015 Messages 941 May 14, 2019 #3 MarkFL said: When you get to: [MATH]r^4=\frac{CL}{R}[/MATH] I would next write: [MATH]r=\pm\sqrt[4]{\frac{CL}{R}}[/MATH] or: [MATH]r=\pm\left(\frac{CL}{R}\right)^{\frac{1}{4}}[/MATH] This is assuming we are only after real solutions. Click to expand... Ok. Nicely done. I know where the power of 1/4 comes from.
MarkFL said: When you get to: [MATH]r^4=\frac{CL}{R}[/MATH] I would next write: [MATH]r=\pm\sqrt[4]{\frac{CL}{R}}[/MATH] or: [MATH]r=\pm\left(\frac{CL}{R}\right)^{\frac{1}{4}}[/MATH] This is assuming we are only after real solutions. Click to expand... Ok. Nicely done. I know where the power of 1/4 comes from.
