Basic logorithmic equation

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• Mar 24th 2010, 01:19 PM
Drozman03
Basic logorithmic equation
Quote:

The half-life of iodine-123 is about 13 hours. You begin with 50 grams of iodine-123.
Write an equation that gives the amount of iodine-123 remaining after t-hours.

It says that the equation ends up being I(t) = 50(.948)^t. In the middle of finding the final equation there is a part that goes like this: when t = 0 I(t) = 25 so 25 = 50B^13 then it goes into the final equation
I(t) = 50(.948)^t.

My question is how do you get .948 from B?
• Mar 24th 2010, 01:25 PM
pickslides
Quote:

Originally Posted by Drozman03
25 = 50B^13

My question is how do you get .948 from B?

$\displaystyle 25 = 50B^{13}$

$\displaystyle 0.5 = B^{13}$

I would use a spreadsheet from here to trial values for $\displaystyle B$
• Mar 24th 2010, 01:30 PM
integral
$\displaystyle 25 = 50b^{13}$

$\displaystyle \frac{1}{2}=b^{13}$

$\displaystyle b^x=c \therefore log_bc=x$
so
$\displaystyle log_b\frac{1}{2}=13$

$\displaystyle \frac{ln(\frac{1}{2})}{ln(b)}=13$
(log base change formula. Log of base b a equals log a divided by log b or ln a divided by ln b (I like ln))

$\displaystyle \frac{ln(\frac{1}{2})}{13}=ln(b)$

$\displaystyle ln\frac{1}{2}=-ln{2}$
because 1/2 is really 2^-1, powers always go before the log.
$\displaystyle b=e\frac{-ln(2)}{13}=.9481$