Logarithmic Graphing Problem
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"A population of an organism grows such that after t hours the number of organisms is N thousand, where N is given by the equation N = A - $8e^{-kt}$
Initially there are 3000 organisms and this number doubles after 5 hours."
Find the value of : i) A
ii) k
This is all I need help for currently, and I have attempted many ways. Firstly, I did a table, so that when t is 1(as it is initial), N is 3000 and then when t is 6 (5 hours later), N is 6000(doubled).
What I did next was to ln the equation, getting ln N = lnA - 8kt. Using simultaneous equations, I have:
ln 3000 = ln A - 8k and ln 6000 = ln A - 48k and this is where I'm stuck, send help D;
logarithms
asked Jan 1 at 14:14
Jia Xuan NgJia Xuan Ng
202
$endgroup$
add a comment |
$begingroup$
"A population of an organism grows such that after t hours the number of organisms is N thousand, where N is given by the equation N = A - $8e^{-kt}$
Initially there are 3000 organisms and this number doubles after 5 hours."
Find the value of : i) A
ii) k
This is all I need help for currently, and I have attempted many ways. Firstly, I did a table, so that when t is 1(as it is initial), N is 3000 and then when t is 6 (5 hours later), N is 6000(doubled).
What I did next was to ln the equation, getting ln N = lnA - 8kt. Using simultaneous equations, I have:
ln 3000 = ln A - 8k and ln 6000 = ln A - 48k and this is where I'm stuck, send help D;
logarithms
asked Jan 1 at 14:14
Jia Xuan NgJia Xuan Ng
202
$endgroup$
$begingroup$
t=0 is the initial time. So A should be easier to calculate. Then you need to solve 2N = A - 8$e^{-5k}$, which should also be doable.
$endgroup$
– Joel Pereira
Jan 1 at 14:23
add a comment |
$begingroup$
"A population of an organism grows such that after t hours the number of organisms is N thousand, where N is given by the equation N = A - $8e^{-kt}$
Initially there are 3000 organisms and this number doubles after 5 hours."
Find the value of : i) A
ii) k
This is all I need help for currently, and I have attempted many ways. Firstly, I did a table, so that when t is 1(as it is initial), N is 3000 and then when t is 6 (5 hours later), N is 6000(doubled).
What I did next was to ln the equation, getting ln N = lnA - 8kt. Using simultaneous equations, I have:
ln 3000 = ln A - 8k and ln 6000 = ln A - 48k and this is where I'm stuck, send help D;
logarithms
asked Jan 1 at 14:14
Jia Xuan NgJia Xuan Ng
202
$endgroup$
"A population of an organism grows such that after t hours the number of organisms is N thousand, where N is given by the equation N = A - $8e^{-kt}$
Initially there are 3000 organisms and this number doubles after 5 hours."
Find the value of : i) A
ii) k
This is all I need help for currently, and I have attempted many ways. Firstly, I did a table, so that when t is 1(as it is initial), N is 3000 and then when t is 6 (5 hours later), N is 6000(doubled).
What I did next was to ln the equation, getting ln N = lnA - 8kt. Using simultaneous equations, I have:
ln 3000 = ln A - 8k and ln 6000 = ln A - 48k and this is where I'm stuck, send help D;
logarithms
logarithms
asked Jan 1 at 14:14
Jia Xuan NgJia Xuan Ng
202
asked Jan 1 at 14:14
Jia Xuan NgJia Xuan Ng
202
asked Jan 1 at 14:14
Jia Xuan NgJia Xuan Ng
202
asked Jan 1 at 14:14
Jia Xuan NgJia Xuan Ng
202
asked Jan 1 at 14:14
Jia Xuan NgJia Xuan Ng
202
202
$begingroup$
t=0 is the initial time. So A should be easier to calculate. Then you need to solve 2N = A - 8$e^{-5k}$, which should also be doable.
$endgroup$
– Joel Pereira
Jan 1 at 14:23
add a comment |
$begingroup$
t=0 is the initial time. So A should be easier to calculate. Then you need to solve 2N = A - 8$e^{-5k}$, which should also be doable.
$endgroup$
– Joel Pereira
Jan 1 at 14:23
$begingroup$
t=0 is the initial time. So A should be easier to calculate. Then you need to solve 2N = A - 8$e^{-5k}$, which should also be doable.
$endgroup$
– Joel Pereira
Jan 1 at 14:23
$begingroup$
t=0 is the initial time. So A should be easier to calculate. Then you need to solve 2N = A - 8$e^{-5k}$, which should also be doable.
$endgroup$
– Joel Pereira
Jan 1 at 14:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First note that N is how many thousand of organisms there are, not how many organisms there are.
At time $t=0$, $N=3$. Plugging this into your equation gives
$$3 = A - 8e^{-0k}$$
$$3 = A - 8$$
$$A = 11$$.
At time $t=5$, $N=6$. Plugging this into your equation gives
$$6 = 11 - 8e^{-5k}$$
$$-5 = -8e^{-5k}$$
$$e^{-5k} = frac58$$
$$-5k = lnfrac58$$
$$k = -frac{1}{5}lnfrac58$$
answered Jan 2 at 0:22
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Jia Xuan Ng
Jan 2 at 8:24
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
First note that N is how many thousand of organisms there are, not how many organisms there are.
At time $t=0$, $N=3$. Plugging this into your equation gives
$$3 = A - 8e^{-0k}$$
$$3 = A - 8$$
$$A = 11$$.
At time $t=5$, $N=6$. Plugging this into your equation gives
$$6 = 11 - 8e^{-5k}$$
$$-5 = -8e^{-5k}$$
$$e^{-5k} = frac58$$
$$-5k = lnfrac58$$
$$k = -frac{1}{5}lnfrac58$$
answered Jan 2 at 0:22
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Jia Xuan Ng
Jan 2 at 8:24
add a comment |
$begingroup$
First note that N is how many thousand of organisms there are, not how many organisms there are.
At time $t=0$, $N=3$. Plugging this into your equation gives
$$3 = A - 8e^{-0k}$$
$$3 = A - 8$$
$$A = 11$$.
At time $t=5$, $N=6$. Plugging this into your equation gives
$$6 = 11 - 8e^{-5k}$$
$$-5 = -8e^{-5k}$$
$$e^{-5k} = frac58$$
$$-5k = lnfrac58$$
$$k = -frac{1}{5}lnfrac58$$
answered Jan 2 at 0:22
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Jia Xuan Ng
Jan 2 at 8:24
add a comment |
$begingroup$
First note that N is how many thousand of organisms there are, not how many organisms there are.
At time $t=0$, $N=3$. Plugging this into your equation gives
$$3 = A - 8e^{-0k}$$
$$3 = A - 8$$
$$A = 11$$.
At time $t=5$, $N=6$. Plugging this into your equation gives
$$6 = 11 - 8e^{-5k}$$
$$-5 = -8e^{-5k}$$
$$e^{-5k} = frac58$$
$$-5k = lnfrac58$$
$$k = -frac{1}{5}lnfrac58$$
answered Jan 2 at 0:22
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
First note that N is how many thousand of organisms there are, not how many organisms there are.
At time $t=0$, $N=3$. Plugging this into your equation gives
$$3 = A - 8e^{-0k}$$
$$3 = A - 8$$
$$A = 11$$.
At time $t=5$, $N=6$. Plugging this into your equation gives
$$6 = 11 - 8e^{-5k}$$
$$-5 = -8e^{-5k}$$
$$e^{-5k} = frac58$$
$$-5k = lnfrac58$$
$$k = -frac{1}{5}lnfrac58$$
answered Jan 2 at 0:22
Erik ParkinsonErik Parkinson
1,17519
answered Jan 2 at 0:22
Erik ParkinsonErik Parkinson
1,17519
answered Jan 2 at 0:22
Erik ParkinsonErik Parkinson
1,17519
answered Jan 2 at 0:22
Erik ParkinsonErik Parkinson
1,17519
1,17519
$begingroup$
Thank you very much!
$endgroup$
– Jia Xuan Ng
Jan 2 at 8:24
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Jia Xuan Ng
Jan 2 at 8:24
$begingroup$
Thank you very much!
$endgroup$
– Jia Xuan Ng
Jan 2 at 8:24
$begingroup$
Thank you very much!
$endgroup$
– Jia Xuan Ng
Jan 2 at 8:24
add a comment |
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$begingroup$
t=0 is the initial time. So A should be easier to calculate. Then you need to solve 2N = A - 8$e^{-5k}$, which should also be doable.
$endgroup$
– Joel Pereira
Jan 1 at 14:23