- WELCOME TO A LESSON ON SOLVING FOR TIME
IN THE ANNUITY AND LOAN FORMULAS USING COMMON LOGARITHMS.
THIS LESSON DOES ASSUME YOU'VE ALREADY WATCHED THE VIDEO
ON HOW TO SOLVE EXPONENTIAL EQUATIONS
USING COMMON LOGARITHMS.
OFTEN WE ARE INTERESTED
IN HOW LONG IT WILL TAKE TO ACCUMULATE MONEY
OR HOW LONG WE NEED TO EXTEND A LOAN
TO BRING PAYMENTS DOWN TO A REASONABLE LEVEL.
THIS REQUIRES SOLVING FOR TIME IN THE ANNUITY OR LOAN FORMULAS,
WHICH WE SEE HERE BELOW.
IN BOTH CASES, CAPITAL N IS TIME IN YEARS.
TO SOLVE FOR N, WE'LL USE LOGARITHMS.
MORE SPECIFICALLY, WE'LL BE USING COMMON LOGARITHMS.
IN THIS LESSON WE'LL BE USING THE ANNUITY FORMULA,
AND THEN WE'LL SHOW HOW TO SET UP A SIMILAR PROBLEM
USING THE LOAN FORMULA.
LET'S BEGIN BY REVIEWING WHAT THE VARIABLES REPRESENT
IN THE ANNUITY FORMULA.
P SUB N IS THE BALANCE IN THE ACCOUNT AFTER N YEARS.
D IS THE REGULAR DEPOSIT OR DEPOSIT EVERY TIME PERIOD.
R IS THE ANNUAL INTEREST RATE AS A DECIMAL,
AND K IS THE NUMBER OF COMPOUNDING PERIODS IN ONE YEAR.
LET'S TAKE A LOOK AT OUR EXAMPLE.
IF YOU INVEST $200 PER MONTH IN AN ACCOUNT
THAT PAYS 3.5% COMPOUNDED MONTHLY INTEREST,
HOW LONG WILL IT TAKE FOR THE ACCOUNT TO GROW TO $12,000?
LET'S BEGIN BY LISTING OUT THE GIVEN INFORMATION.
IF YOU'RE INVESTING $200 PER MONTH, THAT MEANS THAT D IS 200.
THE INTEREST RATE IS 3.5%.
AS A DECIMAL THAT WOULD BE 0.035, WHICH IS R.
INTEREST IS COMPOUNDED MONTHLY,
AND SINCE THERE ARE 12 MONTHS IN A YEAR, K IS 12.
WE WANT TO KNOW HOW LONG IT WILL TAKE FOR THE ACCOUNT
TO GROW TO $12,000.
SO P SUB N WOULD BE 12,000.
WE DON'T KNOW THE TIME.
SO WE'RE TRYING TO SOLVE FOR N,
AND NOW, WE'LL SUB THESE VALUES INTO OUR EQUATION
AND THEN SOLVE FOR N.
SO NOTICE THAT D = 200, K = 12,
WHICH OCCURS HERE, HERE AND HERE.
R = 0.035, WHICH IS HERE AND HERE,
AND P SUB N = 12,000.
SO AGAIN, OUR GOAL HERE IS TO SOLVE THIS FOR N,
AND BECAUSE N IS IN THE EXPONENT,
THIS IS WHY WE NEED LOGARITHMS TO SOLVE THIS EQUATION.
NOW, TO BEGIN, NOTICE HOW WE HAVE THIS FRACTION
HERE IN THE DENOMINATOR,
AND THIS FRACTION BAR MEANS DIVISION.
SO NOTICE HOW IF WE MULTIPLY BOTH SIDES OF THE EQUATION
BY THIS FRACTION HERE,
WE CAN ELIMINATE THIS FRACTION FROM THE RIGHT SIDE.
AGAIN, IN THIS FIRST STEP,
WE MULTIPLY BOTH SIDES OF THE EQUATION BY THIS FRACTION HERE.
IT'S HELPFUL IF WE CAN THINK OF THIS AS BEING OVER ONE,
AND THEREFORE, THESE TWO SIMPLIFY TO ONE,
AND THIS FRACTION x 12,000 = 35.
LET'S JUST VERIFY THAT.
WE HAVE .035 DIVIDED BY 12 AND THEN x 12,000.
SO THAT'S WHERE THE 35 CAME FROM.
NOW, FOR THE NEXT STEP, IT SAYS 200 x THIS QUANTITY HERE.
SO LET'S GO AHEAD AND DIVIDE BOTH SIDES BY 200,
WHICH MEANS THIS WILL SIMPLIFY TO ONE.
SO WE CAN DROP THESE OUTER PARENTHESES,
AND WE'LL LEAVE THIS AS A FRACTION,
EVEN THOUGH WE COULD SIMPLIFY IT.
SO NOW, WE HAVE 35 DIVIDED BY 200 = THIS QUANTITY HERE.
AGAIN, NOTICE HOW THIS IS RAISED TO THE 12N POWER.
SO FOR THE NEXT STEP, WE WILL ADD ONE TO BOTH SIDES.
SO WE'D HAVE 35/200 + 1.
AGAIN, WE COULD TRY TO SIMPLIFY THIS, BUT I'M NOT GOING TO,
AND THIS IS EQUAL TO THE QUANTITY 1 + .035 DIVIDED BY 12
RAISED TO THE POWER OF 12N.
THIS IS WHERE THE COMMON LOG IS GOING TO HELP US SOLVE THIS.
LET'S GO AHEAD AND PUT THIS IN PARENTHESES, TOO.
WE'RE GOING TO TAKE THE COMMON LOG
OF BOTH SIDES OF THE EQUATION.
BY DOING THIS, WE CAN APPLY THE POWER PROPERTY OF LOGARITHMS
HERE, MEANING WE CAN TAKE THE EXPONENT OF 12N
AND MOVE IT TO THE FRONT.
SO WE'D HAVE 12N x THIS LOGARITHM.
SO NOW, WE HAVE LOG OF 35 DIVIDED BY 200 + 1
= 12N x LOG OF 1 + .035 DIVIDED BY 12,
AND NOW, WE CAN ACTUALLY SOLVE FOR N.
WE WANT TO ISOLATE N.
WE DON'T WANT THIS 12 HERE OR THIS LOG,
WHICH MEANS WE CAN NOW DIVIDE BOTH SIDES
BY 12 LOG OF THE QUANTITY 1 + .035 DIVIDED BY 12,
AND THIS WILL GIVE US THE VALUE OF N.
NOTICE IF WE SIMPLIFY THE RIGHT SIDE, 12/12 SIMPLIFIES TO 1,
AND THESE TWO LOGS ARE THE SAME.
SO THIS ALSO SIMPLIFIES TO ONE.
WE COULD HAVE CONVERTED SOME OF THIS TO DECIMALS EARLIER,
BUT THEN IF WE USE ROUNDED DECIMALS
TO PERFORM MORE CALCULATIONS,
OUR ANSWER WOULD NOT BE AS ACCURATE.
SO NOW, ON THE RIGHT SIDE WE JUST HAVE N,
AND TO EVALUATE THE LEFT SIDE, WE'LL GO BACK TO THE CALCULATOR.
WE DO WANT TO PUT THE ENTIRE NUMERATOR AND DENOMINATOR
IN ITS OWN SET OF PARENTHESES.
SO WE'LL HAVE AN OPEN PARENTHESIS COMMON LOG,
WHICH IS HERE, AND THEN WE HAVE 35 DIVIDED BY 200 + 1.
NOW, WE'LL NEED A CLOSE PARENTHESIS FOR THE LOGARITHM
AND ANOTHER CLOSE PARENTHESIS FOR THE NUMERATOR
AND THEN DIVIDED BY OPEN PARENTHESIS FOR THE DENOMINATOR,
12 COMMON LOG 1 + .035 DIVIDED BY 12,
CLOSE PARENTHESIS FOR THE LOGARITHM,
ANOTHER CLOSE PARENTHESIS FOR THE DENOMINATOR AND ENTER.
SO IF WE ROUND TO THE NEAREST TENTH,
WE CAN SAY THIS WOULD TAKE APPROXIMATELY 4.6 YEARS.
SO ONCE AGAIN, THIS MEANS THAT IF YOU INVEST $200 PER MONTH
IN AN ACCOUNT THAT PAYS 3.5% COMPOUNDED MONTHLY INTEREST,
IN APPROXIMATELY 4.6 YEARS,
THE BALANCE WOULD BE APPROXIMATELY $12,000.
NOW, THIS EXAMPLE WAS FOR AN ANNUITY,
BUT THE PROCESS WOULD BE THE SAME IF WE USE A LOAN FORMULA,
EVEN THOUGH THE EQUATION WOULD LOOK A LITTLE BIT DIFFERENT.
SO BEFORE WE GO,
LET'S AT LEAST SET ONE UP THAT INVOLVES A LOAN FORMULA.
FOR EXAMPLE, IF YOU PURCHASE A $799 TABLET ON YOUR CREDIT CARD,
WHICH HAS AN INTEREST RATE OF 14% COMPOUNDED MONTHLY,
HOW LONG WILL IT TAKE FOR YOU PAY OFF THE PURCHASE
IF YOU MAKE PAYMENTS OF $75 A MONTH?
SO USING OUR LOAN FORMULA THIS TIME,
NOTICE THAT P SUB 0 OR THE PRINCIPAL WOULD BE $799.
THE MONTHLY PAYMENT, D, IS $75.
THE INTEREST RATE AS A DECIMAL WOULD BE 0.14,
AND BECAUSE THE INTEREST IS COMPOUNDED MONTHLY,
AND THERE'S 12 MONTHS IN A YEAR, K IS 12.
NOW, WE COULD FOLLOW THE SAME PROCEDURE THAT WE DID
FOR THE ANNUITY FORMULA
AND SOLVE THIS FOR N TO ANSWER THE QUESTION.
I HOPE YOU FOUND THIS HELPFUL.