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anonymous
 one year ago
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
anonymous
 one year ago
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

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DanJS
 one year ago
Best ResponseYou've already chosen the best response.0\[A = A _{0}*e^{r*t}\] you want A to be half the original amount A sub zero. that is the equation they gave you

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0notice now, both sides have an a, divide both sides by a to eliminate that

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 2 }=e^{4.457*t}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now you do loge^1/2=4.457?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0log base e is the natural log, ln you can do that to both sides

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln(\frac{ 1 }{ 2 })=\ln(e^{4.457*t})\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0hmm, do you remember the log exponent rule.. \[\log_{b}(x^a) = a *\log_{b} (x) \]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0and also, \[\log_{b}(b) = 1 \]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0so on the right, pull down the exponent into the front of the log... and replace the natural log of e with 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not good at this at all :(

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln(\frac{ 1 }{ 2 })=\ln(e^{4.457*t}) = 4.457t * \ln(e)\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0just have to remember the rules for logs

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0ln(e) = 1, so ln(1/2) = 4.457*t just have to solve for t, with a calculator

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yep, just round to the thousandths place, 1.555 days

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could you help me with one more please?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0is e^(x+1) all exponent, or just x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0similar to last prob.. take the ln of both sides ln[e^(x+1)] = ln(1.038)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0exponent rule brings (x+1) to the front (x+1) * ln(e) = ln(1.038)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0natural log of e is 1, so x + 1 = ln(1.038) x = ln(1.038) + 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then rounded to the nearest thousandth it would just be 1.037?
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