Basic logorithmic equation

Quote:

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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.

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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?

Quote:

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Originally Posted by Drozman03

25 = 50B^13



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

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I would use a spreadsheet from here to trial values for

so



(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))





because 1/2 is really 2^-1, powers always go before the log.