Question

Stuck with this.. Help me!!

Answer 1

is this correct:
a^10=150000/120000

Answer 2

@Mertsj can u help me with it?

Answer 3

Is the formula
\[A=A _{0}e ^{kt}\]

Answer 4

yea right!

Answer 5

First we have to find the constant.

Answer 6

a^t=y2/y1?

Answer 7

actually the formula we are using is \[N(t)=C _{0} a ^{t}\]

Answer 8

\[150000=120000e ^{10k}\]
\[1.25=e ^{10k}\]
\[\ln 1.25=10k\]
\[k=.022314355\]

Answer 9

Is that what you got for k?

Answer 10

What is C_0 and a in your formula?

Answer 11

ur formula is right also, a is the growth factor and C0 is called the initial value..

Answer 12

\[A=120000e ^{37(.022314355)}\]
A=273999 in the year 2017
Is that what you got?

Answer 13

just give me a sec..

Answer 14

Then I get 1.022565182 for the growth factor.

Answer 15

yea i got the same thing

Answer 16

for part a

Answer 17

for b.. is it like (0.022314355-1) x 100%

Answer 18

\[A=120000(1.022565182)^1=122707.8219\]
\[122708-120000=2708\]
\[\frac{2708}{120000}=2.3 % yearly increase

Answer 19

b is the growth factor.

Answer 20

c is the percent annual increase.

Answer 21

Are we on the same page?

Answer 22

yea..

Answer 23

for c is it (a-1) x 100%?

Answer 24

\[2=1.022565182^{t}\]
\[t=\frac{\ln 2}{\ln 1.022565182}=31 years\]

Answer 25

Yes. It would be just what you said.

Answer 26

okay

Answer 27

If you round it would be 2.3 %, the same thing I got

Answer 28

got it.. i am a little confused how u get the growth factor?

Answer 29

\[1000000=120000(1.022565182)^t\]
\[8.333=(1.022565182)^t\]
\[t=\frac{\ln 8.3333}{\ln 1.022565182}=95\]

Answer 30

95 years after 1980=2075

Answer 31

\[150000=120000a ^{10}\]
Solve that for a

Answer 32

so its like a^10=1.25

Answer 33

1.022...83 it got it now.. so part a and b are kinda same thing!!

Answer 34

\[1.25=a ^{10}\]
\[\ln 1.25=10\ln a\]
\[\frac{\ln 1.25}{10}=\ln a\]
\[.022314355=\ln a\]
\[e ^{.022314355}=a=1.02256518\]

Answer 35

You might say that.

Answer 36

okay..

Answer 37

well thanks for all ur help.. :)

Answer 38

yw