Question

A radio active isotope decays according to the exponential decay equation where t is in days.
Round to the thousandths place.
For the half life: The half life is the solution (t) of the equation : a/2=ae^−4.457t

Answer 1

\[A = A _{0}*e^{r*t}\]

you want A to be half the original amount A sub zero. that is the equation they gave you

Answer 2

notice now, both sides have an a, divide both sides by a to eliminate that

Answer 3

\[\frac{ 1 }{ 2 }=e^{-4.457*t}\]

Answer 4

now you do loge^1/2=-4.457?

Answer 5

log base e is the natural log, ln

you can do that to both sides

Answer 6

\[\ln(\frac{ 1 }{ 2 })=\ln(e^{-4.457*t})\]

Answer 7

0.6931=-1.4944?

Answer 8

hmm, do you remember the log exponent rule..
\[\log_{b}(x^a) = a *\log_{b} (x) \]

Answer 9

and also, \[\log_{b}(b) = 1 \]

Answer 10

so on the right, pull down the exponent into the front of the log... and replace the natural log of e with 1.

Answer 11

im not good at this at all :(

Answer 12

\[\ln(\frac{ 1 }{ 2 })=\ln(e^{-4.457*t}) = -4.457t * \ln(e)\]

Answer 13

just have to remember the rules for logs

Answer 14

ln(e) = 1, so

ln(1/2) = -4.457*t

just have to solve for t, with a calculator

Answer 15

1.5550?

Answer 16

yep, just round to the thousandths place, 1.555 days

Answer 17

Could you help me with one more please?

Answer 18

e^x+1=1.038?

Answer 19

is

e^(x+1) all exponent, or just x

Answer 20

x+1

Answer 21

similar to last prob..
take the ln of both sides

ln[e^(x+1)] = ln(1.038)

Answer 22

exponent rule brings (x+1) to the front

(x+1) * ln(e) = ln(1.038)

Answer 23

natural log of e is 1, so

x + 1 = ln(1.038)

x = ln(1.038) + 1

Answer 24

1.0372?

Answer 25

yep

Answer 26

then rounded to the nearest thousandth it would just be 1.037?

Answer 27

yes

Answer 28

thank you!!!