- YOU DEPOSIT $1,500 INTO A BANK
THAT PAYS 6% CONTINUOUS INTEREST.
WHAT EQUATION CAN BE USED TO DETERMINE THE ACCOUNT BALANCE
AFTER 2 YEARS,
WHAT IS THE ACCOUNT BALANCE AFTER 5 YEARS,
AND WHEN WILL THE ACCOUNT BALANCE REACH $2,500?
SO WE'LL BE USING THE CONTINUOUS INTEREST FORMULA
GIVE HERE BELOW.
IT'S ACTUALLY THE SAME FORMULA
THAT WE USED FOR EXPONENTIAL GROWTH AND DECAY.
IT'S JUST SOME OF THE VARIABLES HAVE CHANGED.
P REPRESENTS THE PRINCIPAL OR STARTING AMOUNT,
R IS THE ANNUAL INTEREST RATE EXPRESSED AS A DECIMAL,
T IS THE TIME IN YEARS,
AND "A" IS THE AMOUNT AFTER 2 YEARS.
SO FOR THIS SITUATION, WE'D HAVE THE EQUATION "A" = P,
THE STARTING AMOUNT OF 1,500,
x E RAISED TO THE POWER OF R x T.
WELL, THE RATE IS 6%.
6% AS A DECIMAL WOULD BE 0.06 x T, THE TIME IN YEARS.
SO HERE'S OUR EQUATION
THAT WILL TELL US THE ACCOUNT BALANCE AFTER 2 YEARS.
NEXT, WE WANT TO KNOW THE ACCOUNT BALANCE AFTER 5 YEARS,
SO WE'LL SUBSTITUTE 5 FOR T.
SO WE CAN SAY "A" OF 5 IS EQUAL TO 1,500 x E
RAISED TO THE POWER OF 0.06 x 5.
AND NOW, WE'LL USE THE CALCULATOR
TO GET A DECIMAL APPROXIMATION FOR THIS VALUE.
IF WE PRESS 2nd NATURAL LOG,
IT BRINGS UP E RAISED TO THE POWER OF,
THEN WE CAN JUST TYPE IN 0.06 x 5, PRESS ENTER,
AND THE ACCOUNT BALANCE IS GOING TO BE APPROXIMATELY
$2,024.79, RUN INTO THE NEAREST CENT.
AND THIS LAST QUESTION, WE WANT TO KNOW
WHEN THE BALANCE WILL REACH $2500 OR $2,500.
SO WE'RE GIVEN "A," WE WANT TO SOLVE THE EQUATION FOR T,
SO WE'LL HAVE AN EXPONENTIAL EQUATION.
WE NEED TO SOLVE THE EQUATION 2,500 = 1,500 x E
RAISED TO THE POWER OF 0.06T.
SO WE WANT TO SOLVE THIS EQUATION FOR T.
SO WE WANT TO ISOLATE THE EXPONENTIAL PART,
SO WE'LL DIVIDE BOTH SIDES BY 1,500.
THIS SIMPLIFIES TO 1.
THIS IS ACTUALLY A REPEATING DECIMAL,
SO WE'LL LEAVE THIS AS A SIMPLIFIED FRACTION.
NOTICE IF WE TAKE 2,500 DIVIDED BY 1,500.
WE REALLY DON'T WANT TO ROUND THIS VALUE.
WE WANT IT TO BE EXACT SO OUR IN MORE PRECISE.
SO IF WE PRESS MATH, ENTER, ENTER, THIS SIMPLIFIES TO 5/3.
SO WE'LL USE 5/3 HERE
INSTEAD OF A DECIMAL = E RAISED TO THE POWER OF 0.06T.
AND NOW TO SOLVE THIS FOR T,
WE'LL TAKE THE NATURAL LOG OF BOTH SIDES OF THE EQUATION.
ONCE WE DO THIS,
WE CAN APPLY THE POWER PROPERTY OF LOGARITHMS HERE
AND MOVE THIS EXPONENT TO THE FRONT.
SO WE'D HAVE NATURAL LOG 5/3 = 0.06T x NATURAL LOG E.
BUT REMEMBER, NATURAL LOG E IS EQUAL TO 1.
MULTIPLYING BY 1 ISN'T GOING TO CHANGE ANYTHING.
SO MOST OF THE TIME, WE WON'T EVEN WRITE THIS.
SO TO SOLVE THIS FOR T, WE'LL DIVIDE BOTH SIDES BY 0.06.
SO ON THE RIGHT SIDE, THIS SIMPLIFIES TO 1T OR JUST T.
AND NOW, WE'LL USE THE CALCULATOR
TO APPROXIMATE THIS QUOTIENT.
SO WE HAVE NATURAL LOG OF 5/3 DIVIDED BY 0.06.
SO IF ROUNDED TO THE NEAREST 10th OF A YEAR,
THIS WOULD BE APPROXIMATELY 8.5 YEARS.
OKAY, WE'LL TAKE A LOOK AT ONE MORE EXAMPLE
IN THE NEXT VIDEO.