- YOU DEPOSIT $5,500 INTO A BANK
THAT PAYS 3% ANNUAL INTEREST COMPOUNDED MONTHLY.
WHAT EQUATION CAN BE USED TO DETERMINE THE ACCOUNT BALANCE
AFTER T YEARS?
WHAT IS THE ACCOUNT BALANCE AFTER TWO YEARS?
AND WHEN WILL THE BALANCE REACH $6,000?
SO WE'LL BE USING THE COMPOUNDED INTEREST FORMULA
GIVEN HERE BELOW
WHERE P REPRESENTS THE PRINCIPLE OR INITIAL INVESTMENT AMOUNT,
R IS THE ANNUAL INTEREST RATE EXPRESSED AS A DECIMAL.
N IS THE NUMBER OF COMPOUNDS PER YEAR.
NOTICE HOW N OCCURS HERE AND HERE IN THE EXPONENT.
T IS THE TIME IN YEARS, AND "A" IS THE AMOUNT AFTER T YEARS.
SO TO FIND THE EQUATION THAT MODELS THIS ACCOUNT BALANCE
WILL HAVE "A" = THE PRINCIPLE,
A STARTING AMOUNT OF $5,500 x THE QUANTITY 1
+ THE ANNUAL INTEREST RATE EXPRESSED AS A DECIMAL.
3% AS A DECIMAL WOULD BE 0.03.
REMEMBER WE DROP THE PERCENT SIGN AND THEN DIVIDE BY 100
OR MOVE THE DECIMAL POINT TO THE LEFT TWO PLACES.
N IS THE NUMBER OF COMPOUNDS PER YEAR.
IT'S COMPOUNDED MONTHLY, AND THERE ARE 12 MONTHS IN A YEAR
SO N IS 12.
SO WE'RE GOING TO RAISE THIS TO THE POWER OF N x T.
SO, AGAIN, N IS 12 AND T IS THE UNKNOWN TIME IN YEARS.
LET'S GO AHEAD AND FIND THIS SUM HERE.
WE WOULD HAVE "A" = 5,500 x 1 + 0.03 DIVIDED BY 12.
SO WE HAVE 1.0025 RAISED TO THE POWER OF 12T.
AND NOW FOR THE SECOND QUESTION,
TO FIND THE ACCOUNT BALANCE AFTER TWO YEARS
WE JUST SUBSTITUTE TWO FOR T.
SO WE COULD SAY "A" OF 2 = 5,500 x 1.0025
RAISED TO THE POWER OF 12 x 2 WHICH OF COURSE WOULD BE 24.
SO NOW WE'LL GO BACK TO THE CALCULATOR
AND WE'LL GO AHEAD AND ROUND THIS TO THE NEAREST CENT.
SO THE ACCOUNT BALANCE WOULD BE $5,839.66.
NOW, FOR THIS LAST QUESTION WE WANT TO KNOW
WHEN THE BALANCE WILL REACH $6,000.
SO THEY'RE GIVING US "A,"
WE WANT TO SOLVE THE EQUATION FOR T.
SO NOW WE'LL BE SOLVING AN EXPONENTIAL EQUATION
IN WHICH WE'LL USE LOGARITHMS.
SO "A" IS GOING TO BE 6,000.
WE WANT TO SOLVE THIS EQUATION FOR T.
SO THE FIRST STEP IS TO ISOLATE THIS EXPONENTIAL PART.
SO WE'LL DIVIDE BOTH SIDES BY 5,500.
THIS SIMPLIFIES TO 1.
6,000 DIVIDED BY 5,500.
NOTICE HOW THIS IS A REPEATING DECIMAL,
SO WE'LL GO AHEAD AND LEAVE THIS AS A SIMPLIFIED FRACTION.
SO I'LL PRESS MATH, ENTER, ENTER.
SO WE'LL LEAVE THE LEFT SIDE OF THE EQUATION AS 12/11.
SO OVER HERE WE'LL HAVE 1.0025 RAISED TO THE 12T POWER,
AND NOW WE'RE GOING TO TAKE THE NATURAL LOG
OF BOTH SIDES OF THE EQUATION.
WHEN WE DO THIS WE CAN APPLY THE POWER OF PROPERTY
OF LOGARITHMS HERE
AND MOVE THIS EXPONENT OF 12T TO THE FRONT.
SO NOW WE HAVE THE EQUATION NATURAL LOG OF 12/11 = 12T
x NATURAL LOG OF 1.0025.
NOW, WE'RE TRYING TO SOLVE THIS EQUATION FOR T
SO WE'RE GOING TO DIVIDE BOTH SIDES OF THE EQUATION BY 12,
AS WELL AS NATURAL LOG 1.0025.
NOTICE ON THE RIGHT SIDE 12/12 SIMPLIFIES TO 1,
AS WELL AS THESE TWO NATURAL LOGS.
SO WE JUST HAVE T ON THE RIGHT SIDE.
SO T IS EQUAL TO THIS QUOTIENT HERE.
WE'LL HAVE TO GET A DECIMAL APPROXIMATION ON THE CALCULATOR.
SO THE NUMERATOR IS NATURAL LOG 12 DIVIDED BY 11
AND THE DENOMINATOR IS GOING TO BE 12 NATURAL LOG 1.0025.
NOTICE HOW WE HAVE A SET OF PARENTHESIS
AROUND THE NUMERATOR AND DENOMINATOR
TO MAKE SURE IT CALCULATES THIS QUOTIENT CORRECTLY.
SO T IS GOING TO BE APPROXIMATELY,
LET'S SAY 2.9 YEARS.
COMPARING THIS TO OUR ANSWER FOR THE SECOND QUESTION,
THIS DOES SEEM LIKE A REASONABLE ANSWER.
OKAY. I HOPE YOU FOUND THIS HELPFUL.