Question

1500=200e^(25r) solve for r

Answer 1

divide by 200,

then take the natural log of both sides

Answer 2

then divide by 25

Answer 3

so log(1500/200)=25r log(e)?

Answer 4

is that right?

Answer 5

natural log of x is \[\log_e(x)=\ln(x)\]

Answer 6

so \[\log_e(e)=\ln (e)=1\]

Answer 7

so log (e) =1?

Answer 8

i dint get it

Answer 9

mind you e is the mathematical constant. i.e. e=2.718....

Answer 10

when you write log it means the log to base 10\[\log (x)=\log_{10}(x)\]
so \[\log (10)=\log_{10}(10)=1\]
___
but in theis question the base is e, so take the natural log instead of the log to base ten

Answer 11

so is hould ln(e)

Answer 12

?

Answer 13

so \[\ln(1500/200)=25r \ln(e)\]

Answer 14

now the ln (e) but simplifies to one

Answer 15

ln(1500/200)= 2.105

ln(1500/200)/ln(e)=2.105?

Answer 16

/?

Answer 17

i put ln(1500/200) into the calculator. it gave me 2.105

Answer 18

then i divided it by ln(e)

Answer 19

oh, i wouldn't make an approximation
\[\ln(1500/200)=\ln(15/2)=\ln(7.5)\]or\[\ln(1500/200)=\ln(15/2)=\ln(15)-\ln(2)\]

Answer 20

yes. and ln(7.5) = 2.104903021

Answer 21

yeah, if you were going to approximate it

Answer 22

ok so ln(7.5)/ln(e)=ln(7.5)

Answer 23

yes

Answer 24

so ln(7.5)/25 is the answer?

Answer 25

= .0805961208

Answer 26

?

Answer 27

yeah that is a rough value for r ,

Answer 28

if i was you i would leave my answer as

r=ln(7.5)/25

Answer 29

ok

Answer 30

see the real question was The count in a bateria culture was 200 after 15 minutes and 1500 after 40 minutes. What was the initial size of the culture? Find the doubling period. Find the population after 120 minutes. When will the population reach 11000.

Answer 31

i got everything but population after 120 mins.

Answer 32

UnkleRhaukus is right. Also,\[\rm \large \ln|e^{blah}|= blah\]

Answer 33

a general equation for growth is\[A(t)=A_0 e^{t/\tau}\]
where A(t) is the amount at time t
A_0 is the initial amount
tau is the time for the amount to increase by a factor of e

\[1500(r)=200e^{25r} \]

Answer 34

ok... so my formula was wrong

Answer 35

your formula was fine

Answer 36

i think i made a typo though

Answer 37

\[A(r)=200e^{25r}\]
to find the amount at r=120 plug in r=120into the RHS

Answer 38

i have to go now ,

Answer 39

so a(r) = 200e^(25*120)>

Answer 40

?

Answer 41

yeah thats it , just simplify (using a calculator), and round down to the nearest whole number to find A at r=120

Answer 42

calculator cannot calculate bcoz of overflow