Question

The half-life of silver-105 is approximately 41.3 days.

Step 1 of 3 : Determine a so that A(t)=A0a^t describes the amount of silver-105 left after t days, where A0 is the amount at time t=0. Round to six decimal places.

a =

Answer 1

is it A(t) = A0a^t = A(0) = A0a^0 ?

Answer 2

@vocaloid

Answer 3

took me a little bit

if the half-life is 41.3 days, that means at t = 41.3, A(t) is 1/2 of A0

A(t) = A0*a^t

(1/2)A0 = A0*a^41.3

dividing both sides by A0

(1/2) = a^41.3

take the logarithm of both sides

log(1/2) = log(a^41.3)

using the power of logarithms, you can bring that 41.3 exponent down in front of the log

log(1/2) = 41.3log(a)

divide both sides by 41.3

log(1/2)/41.3 = log(a)

evaluate log(1/2)/41.3 on the left side, then raise both sides to the power of 10 to eliminate the log on the right side and isolate a

Answer 4

it turn out it was −0.00728886 by evaluating

Answer 5

good

-0.00728886 = log(a)

raising both sides to base 10

10^(-0.00728886) = a = 0.983357, rounded to 6 places

Answer 6

0.983357

Answer 7

yup

Answer 8

do i put it in with 0. or 983357 since it asked for 6 decimal

Answer 9

it's a decimal number so you do have to include the leading decimal

0.983357 would be the proper way to do it

Answer 10

okay i see