Please help me solve this math problem

Half-live of lead is 22 years. How long will it takes for a sample of this substance to decay to 80% of it original amount? A = A0e^kt
the answer is 7.1 years

A = A0e^kt

Ln(A)= Ln(A0)+kt

Ln(A)- Ln(A0) = kt

Ln(A/A0) = kt

Ln(100/80)= (0.693/22)t

Because half life = 0.693/k or k = 0.693/half life

2.303log(5/4) = (0.693/22)t
Bcoz Ln = 2.303log

t = [2.303( log5 - log4 )*22]/0.693 = 7.08 years

The half-life is 22 years. This means that an original amount of 1 (any units will do) will decay to 0.5 after 22years. You use this fact to determine the value of k.

So A(0) = 1

A(t) = 0.5

t = 22

Substituting into the formula,

0.5 = 1.e^(k.22)

or, simply

0.5 = e^22k

In logarithmic form, this equation just means that

22k = ln(0.5)

ln(0.5) = -0.69315

therefore

22k = -0.69315

k = -0.69315 / 22 = - 0.03151

Now you can use this to find the value of t which gives A(t) = 0.8 :

0.8 = 1.e^(-0.03151 t)

or,

ln(0.8) = -0.03151t

So t = ln(0.8) / (-0.03151)

. . . . = -0.22314 / -0.03151

. . . . = 7.082 years.