SOLUTION: A tumor is injected with 0.3 grams of Iodine-125, which has a decay rate of 1.15% per day. To the nearest day, how long will it take for half of the Iodine-125 to decay? Here i

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Question 1129784: A tumor is injected with 0.3 grams of Iodine-125, which has a decay rate of 1.15% per day. To the nearest day, how long will it take for half of the Iodine-125 to decay?
Here is what I have gotten so far:
radioactive decay model:
A = A₀exp(-kt) ----(1)
A₀ of I-125 = 0.3 g
A = 0.3 - (1.15/100)*0.3 g = 0.29655 g
0.29655= 0.3 exp(-k*1)
At this point when I've gone to evaluate, the answer is completely far off. Could someone please assist in explaining the next steps? Thanks
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

A+=+A%5B0%5D%2Ae%5E%28-kt%29+

given:
A%5B0%5D=0.3
Since the decay rate is 1.15% per day, the amount of I-125 left after t+=+1+day is:
A+=+0.3+-+%281.15%2F100%29%2A0.3+g+=+0.3-0.045=+0.29655g

substituting for A, A₀ and t in eq(1) we get:
0.29655+=+0.3%2Ae%5E%28-k%2A1%29+
ln%280.29655%29+=ln%28+0.3%2Ae%5E%28-k%2A1%29+%29
ln%280.29655%29+=ln%28+0.3%29%2Bln%28e%5E%28-k%2A1%29+%29
ln%280.29655%29+-ln%28+0.3%29=ln%28e%5E%28-k%2A1%29+%29
.........ln%28e%29=1
-k=-0.0115666
k=0.0115666

To the nearest day, how long will it take for half of the Iodine-125 to decay?

0.29655%2F+2=+0.3%2Ae%5E%28-0.0115666%2At%29
t=60.9266

t≈ 61 days