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Nithu05
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#1
Hey guys, could you please explain why the solution shows that the time it takes for the value of N to double is constant?
Last edited by Nithu05; 4 weeks ago
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mqb2766
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#2
(Original post by Nithu05)
Hey guys, could you please explain why the solution shows that the time it takes for the value of N to double is constant?
Hey guys, could you please explain why the solution shows that the time it takes for the value of N to double is constant?
e^(t+ln(2)) =e^(ln(2))*e^t = 2*e^t
So no matter what t is, increasing it by ln(2) will double the value of e^t.
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Nithu05
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(Original post by mqb2766)
Thats pretty much the definition of exponential growth.
e^(t+ln(2)) =e^(ln(2))*e^t = 2*e^t
So no matter what t is, increasing it by ln(2) will double the value of e^t.
Thats pretty much the definition of exponential growth.
e^(t+ln(2)) =e^(ln(2))*e^t = 2*e^t
So no matter what t is, increasing it by ln(2) will double the value of e^t.
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mqb2766
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#4
(Original post by Nithu05)
Thank you but where did you get e^(t+ln(2)) from?
Thank you but where did you get e^(t+ln(2)) from?
20e^(0.04t)
So youre answering the question what is the value of T where t->t+T such that the exponetial doubles in value? Doing the "same" if
T = ln(2)/k = 25ln(2)
then adding T to the exponent is equivalent to multiplying the exponential by 2 as
20e^(0.04(t+T)) = e^(0.04*25ln(2)) * 20e^(0.04t) = 2 * 20e^(0.04t)
Last edited by mqb2766; 4 weeks ago
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Nithu05
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#5
(Original post by mqb2766)
Just a simple exponential, so your function is
20e^(0.04t)
So youre answering the question what is the value of T where t->t+T such that the exponetial doubles in value? Doing the "same" if
T = ln(2)/k = 25ln(2)
then adding T to the exponent is equivalent to multiplying the exponential by 2 as
20e^(0.04(t+T)) = e^(0.04*25ln(2)) * 20e^(0.04t) = 2 * 20e^(0.04t)
Just a simple exponential, so your function is
20e^(0.04t)
So youre answering the question what is the value of T where t->t+T such that the exponetial doubles in value? Doing the "same" if
T = ln(2)/k = 25ln(2)
then adding T to the exponent is equivalent to multiplying the exponential by 2 as
20e^(0.04(t+T)) = e^(0.04*25ln(2)) * 20e^(0.04t) = 2 * 20e^(0.04t)
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mqb2766
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#6
(Original post by Nithu05)
Thank you, slightly clearer but what does ln(2)/k represent; is there a specific formula for exponentials? Furthermore, I substitued 25ln2 into the function 20e^(0.04t) and got 40. Is this something along the lines of what you are trying to explain?
Thank you, slightly clearer but what does ln(2)/k represent; is there a specific formula for exponentials? Furthermore, I substitued 25ln2 into the function 20e^(0.04t) and got 40. Is this something along the lines of what you are trying to explain?
2 = e^(0.04 * T) = e^(0.04 * (ln(2)/0.04))
Whenever you increase t by ln(2)/0.04, so
t -> t + ln(2)/0.04
then the corresponding value of the exponential doubles because of the property above. During covid, the cases were doubling every 4 days. Thats exponential growth. How do you imagine exponential growth and what do you think the question is asking?
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Nithu05
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#7
Th
Thank you so much! I understand now 
(Original post by mqb2766)
40 is certainly double 20, so thats what the question is asking - the time to double.
2 = e^(0.04 * T) = e^(0.04 * (ln(2)/0.04))
Whenever you increase t by ln(2)/0.04, so
t -> t + ln(2)/0.04
then the corresponding value of the exponential doubles because of the property above. During covid, the cases were doubling every 4 days. Thats exponential growth. How do you imagine exponential growth and what do you think the question is asking?
40 is certainly double 20, so thats what the question is asking - the time to double.
2 = e^(0.04 * T) = e^(0.04 * (ln(2)/0.04))
Whenever you increase t by ln(2)/0.04, so
t -> t + ln(2)/0.04
then the corresponding value of the exponential doubles because of the property above. During covid, the cases were doubling every 4 days. Thats exponential growth. How do you imagine exponential growth and what do you think the question is asking?
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