Isotope Decay / half-life Suppose you start with 100g of an isotope that has a half life of 17 years. How long before 20g of the isotope are left?
What's the general formula for these problems? 
 A: From the Wikipedia on Half-Life follows: 
$$n\mathrm{,~the~number~of~half~lifes~elapsed}\\
\mathrm{fraction~remaining} = \frac{1}{2^n}$$
A fraction of 20 grams of the initial 100 grams of isotope is 0.2. Therefore, an answer to the following equation is sought: 
$$0.2 = \frac{1}{2^n} \\ {2^n} = \frac{1}{0.2} =  5 \\ n = {^2}\log{5} = \frac{\log 5}{\log 2}$$
With $n$ being 17 years, the answer to your question is: $$17 \times \frac{\log 5}{\log 2} \approx 39.4727776131 \rightarrow 39 ~\mathrm{years}$$
A: Hint:
$N(t)=N_0(\dfrac{1}{2})^{\dfrac{t}{t_{1/2}}}$
Here $N(t)=20g$ , $N_0=100$ and $t_{\frac{1}{2}}=17$years
A: Let $A(t)$ be the amount of the substance at time $t$. The basic formula for $A(t)$ is
$$A(t)=A(0)e^{-\lambda t}\tag{$1$}$$
Here $\lambda$ is a constant for any particular isotope. 
Note that this formula is not absolutely correct. For example, it does not apply when we have a small number of atoms of the substance. At the extreme end, if today we have three atoms of the isotope in a box, then the formula certainly does not apply, except in a more complicated probabilistic sense. However, for ordinary sized quantities, even in the microgram range, the formula is exceedingly accurate.
The parameter $\lambda$ could be used to describe the decay behaviour of the substance. However, a more traditional measure is the half-life $t_h$. By definition, this is the amount of time for an amount $A(0)$ to decay to $\frac{1}{2}A(0)$. Substituting in Formula $(1)$, we obtain
$$\frac{1}{2}A(0)=A(0)e^{-\lambda t_h}.\tag{$2$}$$
Cancel the $A(0)$, and, to avoid minus signs, rewrite $(2)$ as
$$e^{\lambda t_h}=2.$$
Take the natural logarithm ln of both sides. We get $\lambda t_h=\ln 2$, and therefore $\lambda=\dfrac{\ln 2}{t_h}$. Substituting in Formula $(1)$, we get a formula for $A(t)$ in terms of half-life. It is
$$A(t)=A(0)e^{-\frac{(\ln 2)t}{t_h}}.\tag{$3$}$$
We could manipulate further and obtain a general formula for problems of your type. However, I think it is better to write as follows. 
In our particular case, we have $t_h=17$, $A(0)=100$, and $A(t)=20$. We want to find $t$. Using $(3)$, we get
$$100=20e^{-\frac{(\ln 2)t}{17}}.$$
Manipulation yields $e^{\frac{(\ln 2)t}{17}}=5$. Take the ln of both sides. We get
$$t=17\frac{\ln 5}{\ln 2}.$$
Remark: I would take the point of view that the general formula is $A(t)=A(0)e^{-\lambda t}$. Perhaps it is worth remembering that $\lambda=\frac{\ln 2}{t_h}$ for the sake of exam speed. But it is all too easy to remember things wrong.  If repeatedly we need to find $t$, given $t_h$, $A(t)$, and $A(0)$, it may be worthwhile to develop a general formula for $t$. By imitating the reasoning we used for your specific example, we get
$$t=(t_h)\frac{\ln\left(\frac{A(0)}{A(t)}  \right)}{\ln 2}.$$
