## 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

1. 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

2. DanJS

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

3. DanJS

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

4. anonymous

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

5. DanJS

log base e is the natural log, ln you can do that to both sides

6. DanJS

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

7. anonymous

0.6931=-1.4944?

8. DanJS

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

9. DanJS

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

10. DanJS

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

11. anonymous

im not good at this at all :(

12. DanJS

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

13. DanJS

just have to remember the rules for logs

14. DanJS

ln(e) = 1, so ln(1/2) = -4.457*t just have to solve for t, with a calculator

15. anonymous

1.5550?

16. DanJS

yep, just round to the thousandths place, 1.555 days

17. anonymous

Could you help me with one more please?

18. anonymous

e^x+1=1.038?

19. DanJS

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

20. anonymous

x+1

21. DanJS

similar to last prob.. take the ln of both sides ln[e^(x+1)] = ln(1.038)

22. DanJS

exponent rule brings (x+1) to the front (x+1) * ln(e) = ln(1.038)

23. DanJS

natural log of e is 1, so x + 1 = ln(1.038) x = ln(1.038) + 1

24. anonymous

1.0372?

25. DanJS

yep

26. anonymous

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

27. DanJS

yes

28. anonymous

thank you!!!