calling any and all MATH gurus!

T

themacko

Barking at the moon.
I'm sure this is actually quite easy, but I'm pretty rusty on my calc so if someone could hook me up I'd GREATLY appreciate it.

Here is the following equation:

P=e^( [ -ln(2) / H ] * t )

I need to re-work this equation so, given P and t I can find H. I know you probably have to flip the log or something, but like I said it's been a while since I've done this stuff.

Thanks for any help!!!

BTW, I need it by tomorrow morning. Doh! :eek:
 
I'm sure it's been longer since I took a calculus class.

Just curious... would the value of H be equal to....?

H=[-Ln(2)*T] / Ln(P)

Any way for you to test if that's the answer? I wish I hadn't given away my TI89... :)
 
Hey thanks Dusky! There actually isn't a way to check the answer (yet) because it is more-or-less a bonus question for an exam tomorrow.

Anyways, the stuff we're doing is radiometric dating (finding out how old rocks are by calculating their half-life, etc.) The original equation is to calculate the proportion of the original isotope remaining in a rock after (t) years.

P = proportion of orig. isotope
t = time (years)
h = half-life

Thanks again for your help bud.

Scott
 
I really doubt the answer I gave you is correct. But hey, nobody else is giving it a shot, so let me show you what I did...

Your equation:
P=e^( [ -Ln(2) / H ] * t )

Take LN of both sides:
Ln(P)=Ln(e( [ -Ln(2) / H ] * t ) )

Because ln(e^x)=x...
Ln(P)=[ -Ln(2) / H ] * t

Multiply both sides by H:
H*Ln(P)=-Ln(2)*t

Divide both sides by Ln(P):
H=[-Ln(2)*t] / Ln(P)

Given P and t, you shall have H.

Edit: Ignore the text below, as it is unlikely that your calculator would have a log base blah of (bluh), and not an LN button. But proceed, if you must insist.

If for some reason your calculator does not have an Ln button:

Because log base c of x = Ln(x) / Ln(c)...

The answer above can be displayed as:
H=-t*log base P of (2)
 
Wow...and I have a hard time with algebra. It's good to see that $45k spend on my high school education was worth it. :p
 
To help out with solving algebraics/symbolics, consider downloading and installing X48, a Hewlett Packard HP48 calculator emulator.

It's here:

http://ww.versiontracker.com/dyn/moreinfo/macosx/10946

HP specifically designed the calculator to work with "objects" and is REALLY good at solving symbolics.

If you give it the equation:

'i^-i = (e^Pi)^0.5' and solve for Pi, then force a numerical result, you get 3.141596... which is the correct answer. I don't know enough about math to explain the above equation, it was discovered in 1853 by some mathematician.

The HP is for serious math, unfortunately HP stopped making their handhelds two years ago. The calculator ROMS are in the public domain and X48 runs pretty fast on my G3/500.
 
My answer:

H=-ln(2^t/P)

Here's how I got it:

P=e^(t[-ln(2)/H])
lnP=(-tln2)/H
HlnP=-tln2
H=(-tln2)/lnP
H=-ln(2^t/P)

It's been a while since I've done calculus... almost a month, since that's when the AP exam was. I won't be doing any more math until fall, with Calc BC at college.
 
arden...

I know how your t became an exponent [a*Ln(b)=Ln(b^a)], but I don't recall ever seeing that Ln(a)/Ln(b)=Ln(a/b). If the preceding equation is true, then our answers are the same.
 
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