half-life program. need help

What is the half-life (in days) if the decay each day is 2% ?

Also how can I do this

$\begin{array}{rcl} y &=& a \cdot b^t \qquad | \qquad y = \frac{a}{2} \\ \frac{a}{2}&=& a \cdot b^{\frac{h}{1~\text{day}}} \qquad | \qquad h = \text{half-life} \\ \frac{1}{2}&=& b^h \qquad | \qquad \log{} \\ \log{ (\frac{1}{2}) } &=& h\cdot \log{(b)} \\ h &=& \frac{ \log{ (\frac{1}{2}) } } { \log{(b)} } \qquad | \qquad 2~\% \rightarrow 1-\frac{p}{100}=1-\frac{2}{100} = 0.98= b\\ h &=& \frac{ \log{ (\frac{1}{2}) } } { \log{(0.98)} } \\ h &=& \frac{ -0.30102999566 } {-0.00877392431} \\ h &=& 34.3096184915 ~\text{days} \\ \end{array}$

The half-life is 34.3 days

Another way of looking at this is as follows:

If you start with N0, then after 1 day you are left with 0.98*N0

After 2 days you are left with 0.98^2*N0

After 3 days you are left with 0.98^3*N0

...

After t days you are left with 0.98^t*N0

For t to be the half-life you must have 0.98^t*N0 = N0/2

Divide by N0 and take logs of both sides:

ln(0.98^t) = ln(1/2)

t*ln(0.98) = ln(1/2)

t = ln(1/2)/ln(0.98) ≈ 34.3 days

A scientist collects a 20­g sample of radioactive Iodine­131. After a time,

only 5 g of the sample is radioactive. If the half­life of Iodine­131 is 8.07

days, how long did the scientist allow the sample to decay?