anonymous
  • anonymous
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
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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DanJS
  • DanJS
\[A = A _{0}*e^{r*t}\] you want A to be half the original amount A sub zero. that is the equation they gave you
DanJS
  • DanJS
notice now, both sides have an a, divide both sides by a to eliminate that
DanJS
  • DanJS
\[\frac{ 1 }{ 2 }=e^{-4.457*t}\]

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More answers

anonymous
  • anonymous
now you do loge^1/2=-4.457?
DanJS
  • DanJS
log base e is the natural log, ln you can do that to both sides
DanJS
  • DanJS
\[\ln(\frac{ 1 }{ 2 })=\ln(e^{-4.457*t})\]
anonymous
  • anonymous
0.6931=-1.4944?
DanJS
  • DanJS
hmm, do you remember the log exponent rule.. \[\log_{b}(x^a) = a *\log_{b} (x) \]
DanJS
  • DanJS
and also, \[\log_{b}(b) = 1 \]
DanJS
  • DanJS
so on the right, pull down the exponent into the front of the log... and replace the natural log of e with 1.
anonymous
  • anonymous
im not good at this at all :(
DanJS
  • DanJS
\[\ln(\frac{ 1 }{ 2 })=\ln(e^{-4.457*t}) = -4.457t * \ln(e)\]
DanJS
  • DanJS
just have to remember the rules for logs
DanJS
  • DanJS
ln(e) = 1, so ln(1/2) = -4.457*t just have to solve for t, with a calculator
anonymous
  • anonymous
1.5550?
DanJS
  • DanJS
yep, just round to the thousandths place, 1.555 days
anonymous
  • anonymous
Could you help me with one more please?
anonymous
  • anonymous
e^x+1=1.038?
DanJS
  • DanJS
is e^(x+1) all exponent, or just x
anonymous
  • anonymous
x+1
DanJS
  • DanJS
similar to last prob.. take the ln of both sides ln[e^(x+1)] = ln(1.038)
DanJS
  • DanJS
exponent rule brings (x+1) to the front (x+1) * ln(e) = ln(1.038)
DanJS
  • DanJS
natural log of e is 1, so x + 1 = ln(1.038) x = ln(1.038) + 1
anonymous
  • anonymous
1.0372?
DanJS
  • DanJS
yep
anonymous
  • anonymous
then rounded to the nearest thousandth it would just be 1.037?
DanJS
  • DanJS
yes
anonymous
  • anonymous
thank you!!!

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