The number of bacteria present after t minutes is given as B=1000e^kt. Amount of bacteria is 9539 after 4 minutes. Find k.
Please show steps. Oh my goodness. Help?
Please show steps. Oh my goodness. Help?
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just take natural logs ! no need to panic !
9539 = 1000e^4k
e^4k = 9.539
4k = ln 9.539
k = ln 9.539 / 4 = .563847 <---------
9539 = 1000e^4k
e^4k = 9.539
4k = ln 9.539
k = ln 9.539 / 4 = .563847 <---------
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Just insert the values you've been given; B =9539, t = 4
9539 = e^(k*4)
Take natural logs
ln(9539) = k*4
9.1631439371 = k*4
k = 2.29 to 2 dp <<<
9539 = e^(k*4)
Take natural logs
ln(9539) = k*4
9.1631439371 = k*4
k = 2.29 to 2 dp <<<