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- THE HALF LIFE OF PALLADUM 100 IS 4 DAYS.
AFTER 14 DAYS, A SAMPLE OF PLATINUM 100
HAS BEEN REDUCED TO A MASS OF 8 MILLIGRAMS.
WE WANT TO DETERMINE THE INITIAL MASS OF THE SAMPLE
AND ALSO DETERMINE THE MASS 6 WEEKS AFTER THE START DATE.
TO MODEL THE EXPONENTIAL DECAY WE HAVE THE OPTION OF USING
AN EXPONENTIAL FUNCTION IN THIS FORM HERE OR THIS FORM HERE.
THE MAJOR DIFFERENCE IS LOOKING AT THIS FIRST FORM,
THE BASE B IS EQUAL TO 1 + R WHERE R IS THE DECAY RATE
PER UNIT OF TIME,
AND IN THE SECOND FUNCTION
K IS THE CONTINUOS DECAY RATE PER UNIT OF TIME.
BUT SINCE THIS QUESTION DOES NOT ASK US ABOUT DECAY RATE
OR CONTINUOS DECAY RATE
I THINK IT'S GOING TO BE EASIER
IF WE USE THIS FORM OF AN EXPONENTIAL FUNCTION
ESPECIALLY SINCE WE'RE GIVEN THE HALF LIFE.
SO IN THIS FORM "A" IS THE INITIAL AMOUNT
OR THE STARTING AMOUNT
SO IF WE'RE TOLD THE HALF LIFE IS 4 DAYS
AND "A" THE INITIAL AMOUNT,
THIS IS TELLING US THAT P OF 4 MUST EQUAL 1/2 x "A".
AGAIN, THE HALF LIFE IS THE TIME IT TAKES
FOR HALF OF THE MATERIAL TO DECAY,
WHICH MEANS OUR FUNCTION P OF T WOULD BE EQUAL TO "A"
TIMES THE BASE OF 1/2 AND THEN TO DETERMINE THE EXPONENT
AGAIN, WHEN T IS 4,
THIS FUNCTION VALUE HAS TO BE 1/2A
WHICH MEANS THE EXPONENT ON THE BASE OF 1/2 MUST BE 1
WHEN T IS 4.
SO WE CAN'T JUST USE AN EXPONENT OF T,
WE'D HAVE TO USE AN EXPONENT OF T DIVIDED BY 4.
NOTICE HOW WHEN T IS EQUAL TO 4
WE WOULD HAVE 1/2 x "A" TO THE 1st WHICH IS EQUAL TO 1/2A
WHICH HAS TO BE TRUE BASED UPON THE HALF LIFE,
WHICH MEANS WE CAN USE THIS EXPONENTIAL FUNCTION
TO MODEL THE DECAY OF PLATINUM 100.
NEXT, WE'RE TOLD THAT AFTER 14 DAYS
A SAMPLE OF PLATINUM 100 HAD BEEN REDUCED TO A MASS
OF 8 MILLIGRAMS.
SO THEY'RE TELLING US THAT WHEN T EQUALS 14 OR 14 DAYS,
P OF 14 THE AMOUNT REMAINING IS EQUAL TO 8 MILLIGRAMS.
OUR JOB IS TO DETERMINE THE INITIAL MASS
OR THE VALUE OF "A".
SO IF WE PERFORM THESE SUBSTITUTIONS
INTO OUR EXPONENTIAL FUNCTION
WE WOULD HAVE THE EQUATION 8 FOR P OF 14
MUST EQUAL "A" x 1/2 RAISED TO THE POWER OF 14/4
AND NOW WE JUST NEED TO SOLVE THIS FOR "A".
WELL WE SHOULD RECOGNIZE THAT 14/4 IS EQUAL TO 7/2.
LET'S GO AHEAD AND SIMPLIFY THIS.
WELL, TO SOLVE THIS FOR "A"
WE JUST NEED TO DIVIDE BOTH SIDES BY 1/2
RAISED TO THE 7/2 POWER.
NOTICE ON THE RIGHT SIDE THIS SIMPLIFIES TO 1.
SO WE'RE LEFT WITH "A"
AND NOW WE CAN GO TO THE CALCULATOR
TO DETERMINE THIS APPROXIMATE VALUE.
SO WE HAVE 8 DIVIDED BY 1/2 RAISED TO THE POWER OF 7/2.
SO THE INITIAL AMOUNT WAS APPROXIMATELY 90.5 MILLIGRAMS.
THEY ALSO WANT US TO DETERMINE THE MASS AFTER 6 WEEKS.
WE NEED TO BE CAREFUL HERE
BECAUSE T IN OUR FUNCTION REPRESENTS THE NUMBER OF DAYS
AND HERE THEY'RE GIVING US THE TIME IN WEEKS
AND SINCE THERE ARE 7 DAYS IN EACH WEEK AND 6 x 7 = 42,
THEY'RE ASKING US TO DETERMINE THE FUNCTION VALUE WHEN T = 42.
AGAIN, 6 WEEKS EQUALS 42 DAYS.
SO WE HAVE P OF 42 IS GOING TO BE EQUAL
TO THE INITIAL AMOUNT WHICH IS APPROXIMATELY
90.5 x 1/2 RAISED TO THE POWER OF 42 DIVIDED BY 4.
THIS DOES SIMPLIFY TO 21/2 BUT WE'LL GO AHEAD
AND LEAVE IT IN THIS FORM AND GO TO THE CALCULATOR.
SO 90.5 x 1/2 RAISED TO THE POWER OF--
LET'S JUST SAY 21/2.
WE DO NEED THIS EXPONENT IN PARENTHESES.
SO NOTICE HOW THERE'S ONLY APPROXIMALLY .06 MILLIGRAMS
REMAINING AFTER 6 WEEKS OR 42 DAYS.
OKAY, I HOPE YOU FOUND THIS HELPFUL.