Why does $\frac{2P}{P}$ give $2$ and not $P$? The formula for annual compounded interest (according to our lecturer) is  $$2P = P(1 + r)^n.$$
($P$: initial investment, $r$: interest rate, $n$: years).
To solve for $n$, he said the "$P$"s cancel out and you can then transform the equation to
$$n =\frac{\ln(2)}{\ln(1 + r)}.$$
But why is that? If there were an $x$ instead of $P$ (so $2x=x(1+r)^n$), you'd reduce it by dividing through by $x$ (so $2x/x$) which would leave $1x$. So, why is this not happening with $P$, and instead we keep the number $2$ but omit $P$?
 A: $$\frac{2P}{P} = \frac{P + P}{P} = \frac{P}{P} + \frac{P}{P} = 1 + 1 = 2$$
A: Careful, remember that $2P$, or $2x$ for that matter, is shorthand for $2\times P$ or $2\times x$.  So, when you write $$\frac{2x}{x}$$ what you really mean is $$\frac{2\times x}{x}$$ and so the division by $x$ undoes the multiplication by $x$ and you are left with $$\frac{2x}{x} = 2.$$
A: if you start with $P$ capital after $n$ years you have $P(n)=P(1+r)^n$. The question here is thus when you've doubled your starting capital, so when $P(n)=2P$, where $2P$ is short for $2 \times P$ of course. You get the equation (with $\times$) written out
$$2\times P = P \times (1+r)^n$$
The two $P$'s cancel out, just as when you solve $2x=6$ to conclude $x=3$.
so you're left with $2 = (1+r)^n$ which reduces to $\ln(2) = n\ln(1+r)$ and so $n= \frac{\ln(2)}{\ln(1+r)}$ etc.
A: Dividing a product by one of its factors gives the other factor: $(ab):b=a$.  On our case we have
$$(2P):P=2.$$
Similarly, subtracting an addend from a sum  gives the other addend: $(a+b)-b=a$.
