 # Ex: Exponential Growth of Bacteria (Intro Question)

– THE NUMBER OF BACTERIA
IN A CULTURE IS N OF T AFTER T HOURS. SO THE FUNCTION THAT MODELS
THE NUMBER OF BACTERIA, OR THE BACTERIA POPULATION, IS N OF T=500 x E
RAISED TO THE POWER OF 0.15T. WHEN WE HAVE EXPONENTIAL
GROWTH, WE CAN GATHER A LOT
OF INFORMATION FROM THE EXPONENTIAL
GROWTH FUNCTION. IN GENERAL,
IF OUR EXPONENTIAL FUNCTION IS P OF T=P SUB 0
x E RAISED TO THE POWER OF KT, T REPRESENTS THE TIME, K REPRESENTS THE EXPONENTIAL
GROWTH RATE AS A DECIMAL, P SUB 0=THE INITIAL AMOUNT
OR STARTING POPULATION, AND P OF T IS THE AMOUNT
OR POPULATION AFTER TIME T. SO IN THIS CASE,
WITH OUR GIVEN FUNCTION, WHEN THEY ASK US TO FIND
THE GROWTH RATE AS A PERCENT, THEY’RE ASKING US TO DETERMINE
THE VALUE OF K AND THEN CONVERT IT
TO A PERCENTAGE. SO THE GROWTH RATE
IN THIS CASE IS=TO 0.15, WHICH AS A PERCENT
WOULD BE 15%. AND BECAUSE OUR TIME IS T IN
HOURS, THIS IS 15% PER HOUR. NEXT, WE’RE ASKED TO DETERMINE
THE INITIAL POPULATION, OR THE POPULATION
WHEN T IS=TO 0. WELL, THE INITIAL POPULATION
IS P SUB 0, SO LOOKING AT OUR FUNCTION, THE INITIAL POPULATION
WOULD BE 500 BACTERIA. WE SHOULD ALSO BE ABLE
TO MAKE THE CONNECTION THAT WE COULD JUST FIND
N OF 0, LETS GO AHEAD
AND JUST VERIFY THAT. WE WOULD HAVE 500 x E RAISED
TO THE POWER OF 0.15 x 0, WHICH WOULD JUST BE E
TO THE 0. WHEN ANYTHING TO THE 0 POWER
IS=TO 1, SO THIS VERIFIES AGAIN, THE INITIAL POPULATION
IS 500 BACTERIA. AND THEN
FOR THE LAST QUESTION, WE WANT TO KNOW
HOW MANY BACTERIA ARE PRESENT AFTER 12 HOURS. SO THEY ARE TELLING US
THAT T IS=TO 12, SO WE WANT TO FIND N OF 12, WHICH WOULD BE 500 x E RAISED
TO THE POWER OF 0.15 x 12. WE’LL GO AHEAD AND EVALUATE
THIS ON THE CALCULATOR. SO IF 500–IF WE PRESS SECOND
LN OR SECOND NATURAL LOG, IT BRINGS UP E
WITH THE EXPONENT KEY. SO WE JUST TYPE IN 0.15 x 12,
CLOSED PARENTHESIS. IF WE ROUND THIS
TO THE NEAREST BACTERIA, IT WOULD BE APPROXIMATELY
3,025 BACTERIA AFTER 12 HOURS. OKAY, THAT’S GOING TO DO IT
FOR THIS PROBLEM. THANK YOU FOR WATCHING. 