Logarithms and exponential functions

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Shas72
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#1
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#1
The mass m grams of a radio active substance is given by the formula m= m0×e^-kt
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Shas72
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#2
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#2
(Original post by Shas72)
The mass m grams of a radio active substance is given by the formula m= m0×e^-kt
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Notnek
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#3
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#3
What's your question?
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Shas72
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(Original post by Shas72)
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I got the ans for a and b. The question c iam not understanding. So k=0.02 and m0= 50
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Notnek
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(Original post by Shas72)
I got the ans for a and b. The question c iam not understanding. So k=0.02 and m0= 50
Do you have any idea for c? Are you able to write down an equation?
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Shas72
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#6
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(Original post by Sir Cumference)
Do you have any idea for c? Are you able to write down an equation?
Its 3.91
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Shas72
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#7
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(Original post by Sir Cumference)
Do you have any idea for c? Are you able to write down an equation?
So I got ln m= -kt+ln m0
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Shas72
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#8
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#8
(Original post by Sir Cumference)
Do you have any idea for c? Are you able to write down an equation?
Iam not able to do question c which is half life of the radioactive substance
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Notnek
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#9
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(Original post by Shas72)
So I got ln m= -kt+ln m0
You don't need k and m0 anymore because you know their values. So first write out the equation with these values substituted in.

You need to find the time taken to get to half its original mass. What's the original mass? So what's half the original mass?
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Shas72
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#10
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(Original post by Sir Cumference)
You don't need k and m0 anymore because you know their values. So first write out the equation with these values substituted in.

You need to find the time taken to get to half its original mass. What's the original mass? So what's half the original mass?
Is the original mass 40.9
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Notnek
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#11
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(Original post by Shas72)
Is the original mass 40.9
How did you get that?
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Shas72
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(Original post by Sir Cumference)
How did you get that?
From the table of values of t and m
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Notnek
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(Original post by Shas72)
From the table of values of t and m
No that's the mass at time t = 10. The original mass is the time at t = 0.
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Shas72
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#14
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#14
(Original post by Sir Cumference)
No that's the mass at time t = 10. The original mass is the time at t = 0.
So how will I calculate?
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Shas72
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#15
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#15
(Original post by Sir Cumference)
No that's the mass at time t = 10. The original mass is the time at t = 0.
Do I have to use geometric progression
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Shas72
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#16
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#16
(Original post by Sir Cumference)
No that's the mass at time t = 10. The original mass is the time at t = 0.
So m0 is 50
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Notnek
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#17
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#17
(Original post by Shas72)
So how will I calculate?
You now know (after solving part b) that the estimated equation for mass is this

m=50e^{-0.02t}

So what is m when t = 0? That's the original mass.
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Notnek
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#18
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(Original post by Shas72)
So m0 is 50
Yes. And that's the original mass because it's the value of m when t = 0. So what's half of the original mass? Now make an equation and solve it.
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Shas72
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#19
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#19
(Original post by Sir Cumference)
Yes. And that's the original mass because it's the value of m when t = 0. So what's half of the original mass? Now make an equation and solve it.
Half is 25. How do I make an equation?
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Shas72
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#20
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#20
(Original post by Sir Cumference)
Yes. And that's the original mass because it's the value of m when t = 0. So what's half of the original mass? Now make an equation and solve it.
So I substituted these values in the formula given and it is 25=50× e^-0.02t
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