Applications to exponential laws of growth and decay

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Kitobeirens

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i have this question which i cant seem to see where i have gone wrong.

1.The rate in cm^3s^-1, at which air is escaping from a hot air balloonat time t seconds is proportional to the volume of air, v CM^3 in the balloon at that instant. initially V=1000

a) show that v=1000e^-kt , where k is a positive constant. ( I can do this part of the question)

Given that V=500 when t=6

b) show that k=1/6 LN 2

this is the part i cant do, my working goes like this:

1.v=1000e^-kt
2.500=1000e^-6k
3.1/2=e^-6k
4.ln(1/2)=lne^-6k
5.ln(1/2)=-6k
6.-1/6ln(1/2)=k

help is appreciated :)
 
Kitobeirens said:
Here is a photo of the question. the problem i have is with part b (Question 1 btw)

the second photo here is my working out.
http://imgur.com/GmFg8wl
where am i going wrong.
thanks guys
Click to expand...
Your first image isn't showing up for me but assuming you have something like
f(t) = 1000 e-k t
f(6) = 500
what is k, your final answer of
k = \(\displaystyle -\, \frac{1}{6}\, ln(\frac{1}{2})\)
is correct. However the answer can be simplified.

So, what are your thoughts on that? BTW: Thanks for showing your work.
 
Ishuda said:
Your first image isn't showing up for me but assuming you have something like
f(t) = 1000 e-k t
f(6) = 500
what is k, your final answer of
k = \(\displaystyle -\, \frac{1}{6}\, ln(\frac{1}{2})\)
is correct. However the answer can be simplified.

So, what are your thoughts on that? BTW: Thanks for showing your work.
Click to expand...
Ah how do I simplify it to the form 1/6ln2 as that is what part b of the question wants me to do. The first photo us just the question I typed up in my first part of the post
 
Kitobeirens said:
Ah how do I simplify it to the form 1/6ln2...?
Click to expand...
It's not so much "simplifying" as "rearranging in the manner that the book is expecting". To accomplish this, you need to apply some exponent and log rules. You arrived at this point:

. . . . .\(\displaystyle -\dfrac{1}{6}\, \ln\left(\dfrac{1}{2}\right)\, =\, k\)

You know from algebra (here) that the following is true:

. . . . .\(\displaystyle a\, =\, \dfrac{a}{1}\, =\, \left(\dfrac{1}{a}\right)^{-1}\)

From studying logs and their rules (here) that the following is true:

. . . . .\(\displaystyle \ln(a^m)\, =\, m\, \cdot\, \ln(a)\)

The assignment is asking you to determine that the growth/decay constant may be expressed as:

. . . . .\(\displaystyle \dfrac{1}{6}\, \ln(2)\, =\, k\)

And, of course, the product of two negatives is a positive.

Can you see how to apply this information to convert from your result to theirs? ;)
 
stapel said:
It's not so much "simplifying" as "rearranging in the manner that the book is expecting". To accomplish this, you need to apply some exponent and log rules. You arrived at this point:

. . . . .\(\displaystyle -\dfrac{1}{6}\, \ln\left(\dfrac{1}{2}\right)\, =\, k[/t

You know from algebra (here) that the following is true:

. . . . .\(\displaystyle a\, =\, \dfrac{a}{1}\, =\, \left(\dfrac{1}{a}\right)^{-1}\)

From studying logs and their rules (here) that the following is true:

. . . . .\(\displaystyle \ln(a^m)\, =\, m\, \cdot\, \ln(a)\)

The assignment is asking you to determine that the growth/decay constant may be expressed as:

. . . . .\(\displaystyle \dfrac{1}{6}\, \ln(2)\, =\, k\)

And, of course, the product of two negatives is a positive.

Can you see how to apply this information to convert from your result to theirs? ;)\)
Click to expand...
\(\displaystyle
I can't see what to do, I'm kinda going brain dead from all the maths but I feel like I'm completely missing something\)
 
stapel said:
It's not so much "simplifying" as "rearranging in the manner that the book is expecting". To accomplish this, you need to apply some exponent and log rules. You arrived at this point:

. . . . .\(\displaystyle -\dfrac{1}{6}\, \ln\left(\dfrac{1}{2}\right)\, =\, k\)

You know from algebra (here) that the following is true:

. . . . .\(\displaystyle a\, =\, \dfrac{a}{1}\, =\, \left(\dfrac{1}{a}\right)^{-1}\)

From studying logs and their rules (here) that the following is true:

. . . . .\(\displaystyle \ln(a^m)\, =\, m\, \cdot\, \ln(a)\)

The assignment is asking you to determine that the growth/decay constant may be expressed as:

. . . . .\(\displaystyle \dfrac{1}{6}\, \ln(2)\, =\, k\)

And, of course, the product of two negatives is a positive.

Can you see how to apply this information to convert from your result to theirs? ;)
Click to expand...
I've just got it.
-1/6(ln1-ln2)
-1/6(0-ln2)
1/6ln2

Thanks for the help. I forgot ln1 =0
 
Kitobeirens said:
I've just got it.
-1/6(ln1-ln2)
-1/6(0-ln2)
1/6ln2

Thanks for the help. I forgot ln1 =0
Click to expand...
Good job! The steps I'd had in mind were these:

. . . . .\(\displaystyle -\dfrac{1}{6}\, \ln\left(\dfrac{1}{2}\right)\)

Convert the fraction to negative-exponent form:

. . . . .\(\displaystyle -\dfrac{1}{6}\, \ln\left(2^{-1}\right)\)

Convert the power inside the log into a multiplier outside the log:

. . . . .\(\displaystyle \left(-\dfrac{1}{6}\right)\, \left(-1\, \cdot\, \ln(2)\right)\)

Move the "minus" signs around...:

. . . . .\(\displaystyle \left(-\dfrac{1}{6}\right)\, (-1)\, \ln(2)\)

...and "cancel":

. . . . .\(\displaystyle \dfrac{1}{6}\, \ln(2)\)

But either way works! ;)
 
Kitobeirens said:
I've just got it.
-1/6(ln1-ln2)
-1/6(0-ln2)
1/6ln2

Thanks for the help. I forgot ln1 =0
Click to expand...
Now that you have gotten that far, you can proceed to
V(t) = \(\displaystyle 1000\, e^{-\frac{t}{6}\, ln(2)}\)
= \(\displaystyle 1000\, e^{ln(2^{-\frac{t}{6}})}\)
or
\(\displaystyle V(t) = 1000\,\,\, 2^{-\frac{t}{6}} \)
 
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