# How to solve these equations

I'm considering a coursera astronomy course and two of the prerequisites are listed below :

Could provide me with an explanation of how to solve points 2 & 3 above ?

-
1) logarithms 2) Basic equations –  Gautam Shenoy Nov 19 '12 at 11:17
@Gautam, do you think that's helpful? –  Gerry Myerson Nov 19 '12 at 11:17
user, you're not going to get "a familiarity with the rudiments of high-school algebra" from an answer on math.stackexchange. You're going to need to find a refresher or bridging course somewhere, and do that before you try the astronomy course. –  Gerry Myerson Nov 19 '12 at 11:20
There's no need for logarithms in question 1. $-2.3 \times 10^{13} \times .8 \times 10^{-28} = (-2.3 \times .8) \times (10^{13} \times 10^{-28}) = -1.84 \times 10^{-15}$. –  littleO Nov 19 '12 at 11:24
@user470184, Gerry's advice is a sound one, imo: if you're serious about a course in astronomy you're going to need waaaaay more than answers to two questions. Think of taking a remedial course in algebra/geometry/calculus in that college/university –  DonAntonio Nov 19 '12 at 11:33

1. When you compute a product of the form $(a\cdot 10^n)(b\cdot 10^m)$ all you have to do is to compute $ab$ first, and then $10^n\cdot 10^m$ using the property: $10^m\cdot 10^n=10^{m+n}$. If $ab$ is greater than $10$ or less than $1$ then it is recommended to write it as a number $p$ such that $1\leqslant |p|\lt10$ times a power of $10$ and then repeat the same procedure we did. In this particular example: \begin{align}(-2.3\times10^{13})\cdot(0.8\times 10^{-28})&=[(-2.3)\times(0.8)]\times[10^{13}\times 10^{-28}]\\ &=(-1.84)\times(10^{13-28})\\ &=-1.84\times 10^{-15}. \end{align}
2. To prove that $$\rm\frac{2GMr}{D^3}=\frac{Gm}{r^2}\iff D=\left(\frac{2M}m\right)^{1/3}r$$ you really have to be very fluent at algebra. For that particular case, you have to multiply both sides by $\rm D^3$ then regroup all the extra terms in the RHS and then take the cubic root to eliminate that exponential power. If you don't know how to do this then you surely have to revise your algebra before taking any astronomy courses.

As Gerry said: You're going to need to find a refresher or bridging course somewhere, and do that before you try the astronomy course. Or in the case of self-study, go look for some good intros to algebra, precalculus, and calc.

I hope this helps.
Best wishes, $\mathcal H$akim.

-
im taking a course in linear algebra. will this provide a good grounding ? –  blue-sky Mar 28 '14 at 16:03
@user470184 Could you provide me with what will be discussed in that course? –  Hakim Mar 28 '14 at 17:14
I'm not sure but here are course details : MIT 18.06 Linear Algebra, Spring 2005 youtube.com/watch?v=ZK3O402wf1c –  blue-sky Mar 28 '14 at 21:12
@user470184 In addition to that go through Khan|Academy's section on algebra, and precalc first. –  Hakim Mar 28 '14 at 21:32