A certain isotope decays from 80 grams to 2.5 grams in 34 days. What is the half life of this isotope?

Fraction of the isotope left in 34 days = (2.5 g) / (80 g) = 1/32 = (1/2)⁵
Hence, 34 days correspond to 5 half-lives.
Half-life of the isotope = (34 days) / 5 = 6.8 days

Radioactive decay....

Radioactive decay is a first-order process and follows this math:
A = Ao e^-kt ..... where A is "amount" or activity, t is the elapsed time and k is the decay constant, which is related to the half-life, t½. k = ln2 / t½

The equation can also be written as
lnA = -kt + lnAo
ln(A/Ao) = -kt
ln(A/Ao) / -t = k
ln(A/Ao) / -t = ln2 / t½
t½ = -ln2(t) / (ln(A/Ao))
t½ = -0.693(34 day) / (ln(2.5/80))
t½ = 6.80 day

t/th = –log₂(N/N₀)
34/th = –log₂(80/2.5) = 5
th= 34/5 = 6.8 days


N(t) = N₀(1/2)^(t/th)
or N(t) = N₀2^(–t/th)
N₀ is initial amount
N(t) is the amount remaining after time t
th is the half life time, ie, time for half the amount to decay
solving for t or th, t/th = –log₂(N/N₀)

Every half-life, half of the sample remains.

After the first half life, there will be 40 g.

After the second, there will be 20 g

After the third, there will be 10 g

After the fourth, there will be 5 g

After the fifth, there will be 2.5 g

So 5 half-lives have passed over 34 days.

5/34 = how many days in one half-life

EDIT: Bill Russell's answer is better