# Ex: Exponential Function Application with Logarithms

– A TABLET COMPUTER
IS WORTH \$1,500 NEW. IT LOSES 12% OF ITS VALUE
EACH YEAR. THE VALUE CAN BE MODELED BY THE
FUNCTION V OF T EQUALS A TIMES B RAISED TO THE POWER OF T, WHERE V OF T IS THE DOLLAR VALUE AND T IS THE NUMBER OF YEARS
SINCE THE PURCHASE. WE WANT TO WRITE A POSSIBLE
EXPONENTIAL EQUATION FOR THE VALUE. SO OUR FUNCTION WILL BE
IN THE FORM OF V OF T. A REPRESENTS THE INITIAL VALUE
OF THE COMPUTER WHICH WILL BE 1,500 AND WE’RE GOING TO HAVE SUM
BASE B RAISED TO THE POWER OF T. SO LET’S TALK ABOUT
HOW WE’RE GOING TO FIND THE BASE OF THIS EXPONENTIAL FUNCTION. WE KNOW EACH YEAR THE COMPUTER
LOSES 12% OF THE VALUE, WHICH MEANS IT RETAINS
88% OF THE VALUE. SO AFTER THE FIRST YEAR, THE
VALUE WOULD BE 1,500 TIMES 0.88. THIS 0.88 REPRESENTS THE 88%
OF THE VALUE THAT IT RETAINS AFTER THE FIRST YEAR. AFTER THE SECOND YEAR THE COMPUTER LOSES ANOTHER 12%
OF THE VALUE OFF THIS AMOUNT OR RETAINS 88%
OF THIS VALUE HERE. SO AFTER THE SECOND YEAR IT
WOULD BE 1,500 TIMES 0.88 TIMES ANOTHER 0.88. AGAIN, THIS SECOND 0.88
REPRESENTS THE AMOUNT OF VALUE RETAINED AFTER THE SECOND YEAR AND THIS PATTERN
IS GOING TO CONTINUE. AFTER THE THIRD YEAR WE’LL HAVE
1,500 TIMES 0.88 TIMES 0.88 TIMES ANOTHER 0.88. SO THE BASE OF OUR EXPONENTIAL
FUNCTION IS GOING TO BE 0.88 WHICH IS THE PERCENT OF VALUE
THE COMPUTER RETAINS AFTER EACH YEAR. SO TO FIND THE VALUE
AFTER THREE YEARS WE NEED TO DETERMINE VIA 3. SO WE’LL SUBSTITUTE 3 FOR T AND EVALUATE THIS
ON THE CALCULATOR. SO WE’LL HAVE 1,500 TIMES 0.88
RAISED TO THE POWER OF 3. SO THE VALUE OF THE COMPUTER
IS APPROXIMATELY 1,022.21 AFTER THREE YEARS. AND NOW FOR THE LAST QUESTION, WHEN WILL THE VALUE BE HALF
OF ITS ORIGINAL VALUE? WELL,
THE STARTING VALUE IS 1,500 SO HALF OF THIS WOULD BE 750. SO WE WANT TO SET D OF T TO 750
AND THEN SOLVE FOR T. SO WE’LL HAVE 750 EQUALS 1,500
TIMES 0.88 RAISED TO THE POWER OF T. SO WE WANT TO ISOLATE THE EXPONENTIAL PART
OF THIS EQUATION SO WE’LL DIVIDE BOTH SIDES
BY 1,500. THIS WILL BE ONE 750
DIVIDED BY 1,500 EQUALS 1/2 OR 0.5 EQUALS 0.88
RAISED TO THE POWER OF T. AND NOW WE CAN USE LOGARITHMS
TO SOLVE THIS FOR T. IF WE TAKE THE NATURAL LOG
OF BOTH SIDES OF THE EQUATION, ON THE RIGHT SIDE WE CAN USE
THE POWER PROPERTY OF LOGARITHMS TO MOVE THIS T TO THE FRONT, GIVING US NATURAL LOG OF 0.5
EQUALS T TIMES NATURAL LOG 0.88. NOW, TO SOLVE THIS EQUATION
FOR T WE CAN DIVIDE BOTH SIDES
BY NATURAL LOG 0.88. NOTICE HERE THIS SIMPLIFIES TO A
1 SO THIS QUOTIENT WILL EQUAL T. SO WE’LL GO BACK
TO THE CALCULATOR. NATURAL LOG 0.5
DIVIDED BY NATURAL LOG 0.88 WILL GIVE US THE VALUE OF T. WHEN ROUNDED TO THE NEAREST 10TH THIS WILL BE APPROXIMATELY
5.4 YEARS. SO AGAIN THE VALUE WILL BE HALF
OF THE ORIGINAL VALUE AFTER 5.4 YEARS OR APPROXIMATELY 5.4 YEARS.