How to deal with the different powers of the same variable here?

Question

In this solution they take out the ro to put it on one side of the equation and give it power 6

How do we take out ro as a common when the two ro's have different powers?

Thank you.

Answer 1

This is the one step after another calculation:

$$ 6Ar_0^{-7} -12Br_0^{-13} = 0$$

$$ 6Ar_0^{-7} = 12Br_0^{-13} $$

$$ Ar_0^{-7} = 2Br_0^{-13} $$

$$ A = 2B*\frac{r_0^{-13}}{r_0^{-7}} $$

$$ A = 2B*(r_0^{-6}) $$

$$ Ar_0^{6} = 2B $$

$$ r_0^{6} = 2B/A $$

$$ r_0 = \sqrt[6](2B/A) $$

And here is an example since the mechanics seems to be unclear:

$$ A*2^{-3} = B*2^{-2} $$

$$ A = B*\frac{2^{-2}}{2^{-3}} $$

$$ A = B*{2^{-2}}*{2^{3}} $$

$$ A = B*2 $$

Answer 2

Factorize: $$ 0 = 6Ar_0^{-7} - 12Br_0^{-13} = 6r_0^{-13} ( Ar_0^6 - 2B ) $$ so $Ar_0^6 = 2B$. I think you can take it from there.

Answer 3

We can write the equation as $$r_0^{-1}(6Ar_0^{-6}-12Br_0^{-12})=0$$ Multiplying both sides by $r_0$ gives $$6Ar_0^{-6}-12Br_0^{-12}=0$$ Now substituting $x=r_0^{-6}$ (notice that $x\neq 0$) yields $$6Ax-12Bx^2=0$$ This we can easily solve, and since $x\neq 0$, we know $x=\frac{6A}{12B}=\frac{A}{2B}$. Now trace back to $r_0$ to obtain $r_0=x^{-1/6}=\left(\frac{A}{2B}\right)^{-1/6}$, so $$r_0=\left(\frac{2B}{A}\right)^{1/6}$$

Hope this helped!