- A BACTERIA CULTURE STARTING WITH 200 BACTERIA
GROWS AT A RATE PROPORTIONAL TO ITS SIZE.
AFTER THREE HOURS, THERE WILL BE 900 BACTERIA.
THIS IS AN EXAMPLE OF AN EXPONENTIAL GROWTH.
THIS IS AN EXPONENTIAL GROWTH MODEL.
SO WE'LL BE USING THE FUNCTION P OF T = P SUB 0 x E
RAISED TO THE POWER OF KT TO MODEL THIS SITUATION,
WHERE P SUB 0 IS A STARTING POPULATION,
K IS THE EXPONENTIAL GROWTH RATE,
T IS THE TIME IN HOURS,
AND P OF T WOULD BE THE POPULATION AFTER T HOURS.
WE FIRST WANT TO EXPRESS THE POPULATION AFTER T HOURS
AS A FUNCTION OF T.
THEN WE WANT TO FIND THE POPULATION AFTER SIX HOURS.
THEN WE WANT TO KNOW WHEN THE POPULATION WILL REACH 5,000.
SO LETS START BY FINDING THE EXPONENTIAL FUNCTION
THAT WILL MODEL THE BACTERIA POPULATION.
WE'RE STARTING WITH 200 BACTERIA,
WHICH MEANS P SUB 0 = 200
AND THEN AFTER THREE HOURS THERE WILL BE 900 BACTERIA.
SO WHEN T = 3 THE POPULATION, OR P OF T,
IS GOING TO BE EQUAL TO 900.
WE'RE GOING TO SUB THESE VALUES INTO OUR EXPONENTIAL FUNCTION
AND THEN SOLVE FOR K, THE EXPONENTIAL GROWTH RATE.
SO WE WANT TO SOLVE THE EQUATION 900 = 200 x E
RAISED TO THE POWER OF K x T,
BUT T IS 3, SO WE HAVE E RAISED TO THE POWER OF 3K.
AND NOW TO SOLVE THIS EXPONENTIAL FUNCTION
WE'LL START BY ISOLATING THE EXPONENTIAL PART,
OR E RAISED TO THE POWER OF 3K,
SO WE'LL DIVIDE BOTH SIDES BY 200.
THIS SIMPLIFIES TO 1.
900 DIVIDED BY 200 IS 4.5,
SO WE HAVE 4.5 = E RAISED TO THE POWER OF 3K.
AND NOW TO SOLVE FOR K
WE'RE GOING TO TAKE THE NATURAL LOG
OF BOTH SIDES OF THE EQUATION.
AFTER DOING THIS WE CAN APPLY THE POWER PROPERTY OF LOGARITHMS
AND MOVE THIS 3K TO THE FRONT OF THE NATURAL LOG.
SO NOW WE HAVE THE EQUATION
NATURAL LOG 4.5 = 3K x NATURAL LOG E.
BUT REMEMBER NATURAL LOG IS LOG BASE E.
SO WE HAVE NATURAL LOG, OR LOG BASE E OF E,
AND SINCE E TO THE 1st POWER IS EQUAL TO E--
AND, AGAIN, REMEMBER IN EXPONENTIAL FORM
WE WORK OUR WAY AROUND THE EQUAL SIGN.
E TO THE 1st = E,
THIS SIMPLIFIES TO 1
AND MULTIPLYING BY 1 DOESN'T CHANGE ANYTHING,
SO WE JUST HAVE THE EQUATION NATURAL LOG 4.5 = 3K.
SO TO SOLVE FOR K WE DIVIDE BOTH SIDES BY 3.
THIS QUOTIENT WILL GIVE US OUR EXPONENTIAL GROWTH RATE.
SO NOW WE'LL GO TO THE CALCULATOR.
NATURAL LOG 4.5 DIVIDED BY 3.
NOW, THE MORE DECIMAL PLACES WE USE,
THE MORE ACCURATE OUR ANSWER IS GOING TO BE,
SO WE'LL USE SIX DECIMAL PLACES.
SO IT'S GOING TO BE APPROXIMATELY 0.501359.
AND NOW THAT WE HAVE OUR EXPONENTIAL GROWTH RATE,
WE KNOW OUR EXPONENTIAL MODEL
IS GOING TO BE P OF T = THE INITIAL POPULATION OF 200
x E RAISED TO THE POWER OF KT.
AND NOW WE KNOW THE VALUE OF K,
SO IT'S GOING TO BE 0.501359 x T,
WHERE T IS THE TIME IN HOURS.
SO LET'S TAKE THIS FUNCTION BACK TO THE PREVIOUS SCREEN
AND ANSWER THE NEXT TWO QUESTIONS.
SO IF WE WANT TO KNOW THE POPULATION AFTER SIX HOURS,
WE WANT TO FIND P OF 6 OR SUBSTITUTE 6 FOR T.
SO P OF 6 WOULD BE THE POPULATION AFTER SIX HOURS,
SO WE WOULD HAVE 200 x E RAISED TO THE POWER OF 0.501359 x 6.
SO NOW WE'LL GO BACK TO THE CALCULATOR
AND TYPE IN THE RIGHT SIDE OF OUR FUNCTION.
WE PRESS SECOND NATURAL LOG, THAT BRINGS UP E
RAISED TO THE POWER OF
AND THE EXPONENT IS OUR EXPONENTIAL GROWTH RATE x 6.
IF WE ROUND TO THE NEAREST BACTERIA,
THIS WOULD ROUND TO 4,050 BACTERIA.
AND THEN FOR OUR LAST QUESTION WE WANT TO KNOW
WHEN THE POPULATION IS GOING TO REACH 5,000.
WELL, WE CAN TELL FROM THE PREVIOUS QUESTION
IT'S GOING TO BE MORE THEN SIX HOURS.
SO, ACTUALLY, IN THIS QUESTION
THEY'RE GIVING US P OF T FOR OUR FUNCTION,
WE WANT TO SOLVE FOR T.
SO WE WANT TO SOLVE THE EQUATION 5,000 = 200 x E
RAISED TO THE POWER OF 0.501359T.
AND TO SOLVE THIS EXPONENTIAL FUNCTION,
WE'LL ISOLATE THE EXPONENTIAL PART,
SO WE'LL START BY DIVIDING BY 200.
SIMPLIFIES TO 1. 5,000 DIVIDED BY 200 = 25.
WE'LL NOW TAKE THE NATURAL LOG OF BOTH SIDES OF THE EQUATION
AND THEN ON THE RIGHT SIDE
WE'LL APPLY THE POWER PROPERTY OF LOGARITHMS,
SO WE'LL MOVE THIS EXPONENT TO THE FRONT.
SO NOW WE'LL HAVE NATURAL LOG 25 = 0.501359T x NATURAL LOG E.
BUT REMEMBER NATURAL LOG E IS = 1,
SO MULTIPLYING BY 1 IS NOT GOING TO CHANGE THE VALUE OF THIS,
SO WE CAN SIMPLIFY THIS OUT.
SO TO SOLVE FOR T WE'LL DIVIDE BOTH SIDES
BY THIS DECIMAL COEFFICIENT.
NOTICE ON THE RIGHT SIDE THIS SIMPLIFIES TO 1,
SO WE JUST HAVE T
AND THEN WE'LL GET A DECIMAL APPROXIMATION FOR THIS QUOTIENT.
SO WE HAVE NATURAL LOG 25 DIVIDED BY 0.501359.
SO WE CAN SEE IT'S GOING TO BE APPROXIMATELY,
LET'S SAY, 6.42 HOURS.
AND THAT'S GOING TO DO IT FOR THIS EXAMPLE.
I HOPE YOU FOUND THIS HELPFUL.