I have two here.
a=b(e^(-c))
solve for c
For that one, I have looked around and tried to figure it out and all i can think is that I need to use Log to solve it.
a=b/(c^2)
explain what happens to the value of a if c is doubled.
For the second one, can take the sign of proportionality.
$\displaystyle a = \frac{b}{c^2}$
$\displaystyle a\ \alpha\ \frac{1}{c^2}$
If c doubles, then;
So, if a becomes a' and c becomes c', it becomes:
$\displaystyle a'\ \alpha\ \frac{1}{c'^2}$
Now, we know that c' = 2c.
$\displaystyle a'\ \alpha\ \frac{1}{(2c)^2}$
$\displaystyle a'\ \alpha\ \frac{1}{4c^2}$
$\displaystyle 4a'\ \alpha\ \frac{1}{c^2}$
So, a = 4a' and hence, a/4 = a'. You conclude that the new a' is quadrupled when compared to the previous a
(here, c' means the new value of c, not the derivative of c.)
[Note, until I find the sign for the proportionality, I'll use alpha instead]