Chemistry Problem Help Please!?

Basic integrated rate law
A = Ao e^(-kt)
Rewritten in terms of natural logs
lnA - lnAo = -kt

Solve for the decay constant, k. The decay constant is directly related to the half-life.

k = (lnA - lnAo) / -t
k = (ln 6.25 - ln 100) / -21hr
k = 0.132 hr^-1

Decay constant and half-life are related by...
t½ = ln2 / k
t½ = 0.693 / 0.132 hr^-1
t½ = 5.25 hr

This solution is the only one based on the first-order decay equations which govern radioactive decay. Electron's method, while it gets the "right" answer would be useless if 6.25% of the remaining radionuclide weren't conveniently attained in 4 half-lives.

The equations, which I have shown you will work for ANY situation involving radioactive decay.
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Let’s determine the fraction that represents 6.25%. To do this, divide 6.25 by 100
6.25/100 to reduce to the lowest terms, divide the numerator and denominator by 6.25.
The fraction is 1/16.

During one half life, half of the isotope decays. This leaves one half of the original amount. During the 2nd half life one half this amount decays. This leaves one fourth of the original amount. If we keep doing this process, when will the fraction be 1/16?

½ , 1/4 , 1/8, 1/16 This 4 half lifes.
To determine the number of years, divide by 4.
Half life = 5.25 hours
Or
Use the following equation.

½ ^n = 1/16
Take log of both sides.
n * log ½ = log 1/16
n = log 1/16 ÷ log ½ = 4
Half life = 21/4 = 5.25 hours

let n be no. of half life passed.

2^n = 0.0625

n log2 = log0.0625

n=-4.

ie 6.25% = 100%/32 = 100% / (2x2x2x2)

ie 21hours = 4half life.

half life = 21/4 = 5.25 hours

Hope it helps:)