1. ## Basic logorithmic equation

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?

2. Originally Posted by Drozman03
25 = 50B^13

My question is how do you get .948 from B?
$25 = 50B^{13}$

$0.5 = B^{13}$

I would use a spreadsheet from here to trial values for $B$

3. $25 = 50b^{13}$

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

$

b^x=c \therefore log_bc=x$

so
$
log_b\frac{1}{2}=13$

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

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

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