Homework Help

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Assasinof6
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Homework Help

Postby Assasinof6 » Wed Sep 22, 2010 9:58 pm UTC

I have determined a half-life function to be P(t) = 100((1/2)^(t/5730)), where P(t) is the percentage of C-14 remaining, and "t" is the elapsed time.

Now, I am given the value P(t)=70, and am expected to solve for t.
Have I properly converted this?

70 = 100(1/2)^(t/5730))
(t/5730) = log(100(1/2))70
t = 5730(log(100(1/2))70)

Also, my calculator doesn't properly convert logarithms. My guess is it already has an assigned base value. How would I change this?
The value I got when doing this was:
~681456.9

It should be around 3000...
I see.
What?
That you don't.

Syrin
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Re: Homework Help

Postby Syrin » Wed Sep 22, 2010 10:03 pm UTC

70 = 100(1/2)^(t/5730))
(t/5730) = log(100(1/2))70

The second line doesn't follow from the first.

++$_
Mo' Money
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Re: Homework Help

Postby ++$_ » Wed Sep 22, 2010 10:06 pm UTC

70 = 100(1/2)^(t/5730))
(t/5730) = log(100(1/2))70
The problems are between these steps. I'm not sure how you got from one to the other. First of all, things are shuffling around between both sides of the equation like nobody's business. Second, it is not true that [imath]\log(ab^n) = n\log(ab)[/imath].

To convert from one base of logarithm to another, just use the change of base formula:
[math]\log_b a = {\log_c a \over \log_c b}[/math]
Use your calculator's log in place of [imath]\log_c[/imath].

EDIT: I should note that your answer is wrong for reasons unrelated to the base that your calculator uses. I just thought I'd explain anyway.

Assasinof6
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Re: Homework Help

Postby Assasinof6 » Wed Sep 22, 2010 10:30 pm UTC

It's not so much that I am have multiple bases, it's just that I don't know how to go from A to B- in this case, isolating "t" in 70 = 100((1/2)^(t/5730))

Could someone show me how to isolate t?
I see.
What?
That you don't.

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Dason
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Re: Homework Help

Postby Dason » Wed Sep 22, 2010 10:42 pm UTC

Assasinof6 wrote:Could someone show me how to isolate t?

Would you mind showing your work between the two steps people have commented on? If we see where exactly you're going wrong we can help you correct it.
double epsilon = -.0000001;

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The Scyphozoa
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Re: Homework Help

Postby The Scyphozoa » Sat Sep 25, 2010 3:54 am UTC

70 = 100(1/2)^(t/5730))
(t/5730) = log(100(1/2))70

This is correct, it's just that it looks like multiplication when it's actually defining the base of the log.

(t/5730) = log{100*1/2}70
Imagine 100*1/2 to be a subscript. Or imagine
(t/5730) = (log 70)/(log 100*1/2)

And yes, to change the base of your log--example: you want to do log base b of a--you do log a / log b. If there is no log{b}a button, the base is probably 10.

I'm actually surprised how often I see people on these forums unable to sort through minor notation errors. Like one instance in the math jokes thread, where someone was trying to "prove" that studying is evil. It came to a part where the value (the word) "absolute" was being multiplied by two other "variables" on each side of the equation, and most people couldn't grasp that "absolute" was a variable and didn't mean |word|.
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Syrin
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Re: Homework Help

Postby Syrin » Sat Sep 25, 2010 4:28 am UTC

The Scyphozoa wrote:
70 = 100(1/2)^(t/5730))
(t/5730) = log(100(1/2))70

This is correct, it's just that it looks like multiplication when it's actually defining the base of the log.

(t/5730) = log{100*1/2}70
Imagine 100*1/2 to be a subscript. Or imagine
(t/5730) = (log 70)/(log 100*1/2)

And yes, to change the base of your log--example: you want to do log base b of a--you do log a / log b. If there is no log{b}a button, the base is probably 10.

I'm actually surprised how often I see people on these forums unable to sort through minor notation errors. Like one instance in the math jokes thread, where someone was trying to "prove" that studying is evil. It came to a part where the value (the word) "absolute" was being multiplied by two other "variables" on each side of the equation, and most people couldn't grasp that "absolute" was a variable and didn't mean |word|.



Well, before you act all condescending - you're still wrong. It would still not work like that. There is a distinct difference between 100(1/2)^(t/5730) and (100(1/2))^(t/5730).

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The Scyphozoa
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Re: Homework Help

Postby The Scyphozoa » Sat Sep 25, 2010 5:24 am UTC

Syrin wrote:
The Scyphozoa wrote:
70 = 100(1/2)^(t/5730))
(t/5730) = log(100(1/2))70

This is correct, it's just that it looks like multiplication when it's actually defining the base of the log.

(t/5730) = log{100*1/2}70
Imagine 100*1/2 to be a subscript. Or imagine
(t/5730) = (log 70)/(log 100*1/2)

And yes, to change the base of your log--example: you want to do log base b of a--you do log a / log b. If there is no log{b}a button, the base is probably 10.

I'm actually surprised how often I see people on these forums unable to sort through minor notation errors. Like one instance in the math jokes thread, where someone was trying to "prove" that studying is evil. It came to a part where the value (the word) "absolute" was being multiplied by two other "variables" on each side of the equation, and most people couldn't grasp that "absolute" was a variable and didn't mean |word|.



Well, before you act all condescending - you're still wrong. It would still not work like that. There is a distinct difference between 100(1/2)^(t/5730) and (100(1/2))^(t/5730).

Oooh... you're right, I didn't catch that. Thanks.

So in that case it would be
7/10=(1/2)^(t/5730)
log{1/2}(7/10)=(t/5730)
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3rdtry wrote:If there ever is another World War, I hope they at least have the decency to call it "World War 2: Episode One"

doogly wrote:murder is a subset of being mean


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