- 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.
