MHB Im back with change subject of formula

[fordy2707]

so I've been taught basic rearranging of formula which I've found easy now I am given questions which are obviously a lot harder and I've entered the mind black hole if you could point me in the right direction with this where I go wrong

make L the subject

$f=\frac{1}{2\pi}$ $\sqrt{\frac{1}{LC}}$

=${f}^{2}=\frac{1}{{2\pi}^{2}}\frac{1}{LC}$

=${f}^{2}=\frac{1}{{2\pi}^{2}LC}$

above I may have already gone wrong but in my head its ok,but here on I am quite unsure if I am allowed to move the formula as I have

=${2\pi}^{2}C{f}^{2}=\frac{1}{L}$ ?? is that correct ?

then I am at a total loss as what to do with that fraction left over in my guess I am wanting make a new fraction and throw the 1 under the rest to make

L=$\frac{{2\pi}^{2}C{F}^{2}}{1}$ but this is just guess territory now as I've not been shown this type of question

 

[kaliprasad]

so I've been taught basic rearranging of formula which I've found easy now I am given questions which are obviously a lot harder and I've entered the mind black hole if you could point me in the right direction with this where I go wrong

make L the subject

$f=\frac{1}{2\pi}$ $\sqrt{\frac{1}{LC}}$

=${f}^{2}=\frac{1}{{2\pi}^{2}}\frac{1}{LC}$

=${f}^{2}=\frac{1}{{2\pi}^{2}LC}$

above I may have already gone wrong but in my head its ok,but here on I am quite unsure if I am allowed to move the formula as I have

=${2\pi}^{2}C{f}^{2}=\frac{1}{L}$ ?? is that correct ?

then I am at a total loss as what to do with that fraction left over in my guess I am wanting make a new fraction and throw the 1 under the rest to make

L=$\frac{{2\pi}^{2}C{F}^{2}}{1}$ but this is just guess territory now as I've not been shown this type of question

There are 2 mistakes ( one is because of latex)
There are
1) ${f}^{2}=\frac{1}{(2\pi)^{2}LC}$ you should put $2\pi$ in () and not in $\{\}$ because $2\pi$ should get squared
2) $L=\frac{1}{(2\pi)^{2}C{f}^{2}}$

 

[fordy2707]

I see, so i treated $2\pi$ as 1 value where its actually 2 different values to be multiplied together .

thanks for your help,im happy with how close I got.i will write your advice into my learning material and should be able to do a sum like that on my own next time

 

[I like Serena]

The "trick" is, when in doubt, to apply basic operations to the left and to the right of the equal sign.
That is, multiply left and right by the same expression:

$$f^2=\frac 1{(2\pi)^2LC} \\
\Rightarrow f^2 \cdot (2\pi)^2C =\frac 1{(2\pi)^2LC} \cdot (2\pi)^2C$$

After that we can simplify the fraction by canceling common factors:
$$\Rightarrow f^2 \cdot (2\pi)^2C =\frac 1{\cancel{(2\pi)^2}L\cancel C} \cdot \cancel{(2\pi)^2}\cancel C \\
\Rightarrow(2\pi)^2f^2C =\frac 1{L}$$

Now we take the inverse $\frac 1 x$ left and right, to get:
$$\Rightarrow \frac{1}{(2\pi)^2f^2C} = L \\
\Rightarrow L = \frac{1}{(2\pi)^2f^2C}
$$