- WE WANT TO SOLVE THE GIVEN EXPONENTIAL EQUATION.
NOTICE, IN THIS EQUATION,
THERE ARE TWO EXPONENTIAL PARTS.
WHEN WE HAVE TWO EXPONENTIAL PARTS,
WE WANT TO PUT ONE EXPONENTIAL PART
ON EACH SIDE OF THE EQUATION, AS GIVEN HERE.
ONCE IT'S IN THIS FORM,
WE'LL TAKE THE NATURAL LOG OF BOTH SIDES OF THE EQUATION
AND THEN SOLVE FOR X.
AND BECAUSE WE'LL BE USING THE CALCULATOR TO SOLVE THIS,
WE WANT TO EITHER USE NATURAL LOG OR COMMON LOG.
LET'S GO AHEAD AND TAKE THE NATURAL LOG
OF BOTH SIDES OF THE EQUATION.
SO, WE'LL HAVE THE NATURAL LOG OF 3
RAISED TO THE POWER OF 2X MINUS 3,
EQUALS THE NATURAL LOG OF 5
RAISED TO THE POWER OF 4X PLUS 5.
NOW FROM HERE,
WE CAN APPLY THE POWER PROPERTY OF LOG RHYTHMS,
WHICH ALLOWS US TO TAKE THIS EXPONENT HERE
AND MOVE IT TO THE FRONT.
SO, WE'LL DO THAT HERE, AS WELL AS HERE.
NOW, WE WHEN REWRITE THE EQUATION
AS THE QUANTITY 2X MINUS 3,
TIMES NATURAL LOG 3 EQUALS THE QUANTITY 4X
PLUS 5 TIMES NATURAL LOG 5.
NOW JUST LIKE ANY OTHER EQUATION,
WE'RE GOING TO CLEAR THE PARENTHESES
AND THEN SOLVE FOR X.
SO, WE'LL DISTRIBUTE NATURAL LOG 3 HERE
AND WE'LL DISTRIBUTE NATURAL LOG 5 HERE.
SO, WE'LL HAVE 2X NATURAL LOG 3,
MINUS 3 NATURAL LOG 3 EQUALS 4X NATURAL LOG 5
PLUS 5 NATURAL LOG 5.
NOW, WE KNOW OUR GOAL IS TO SOLVE FOR X
AND THERE ARE TWO TERMS THAT HAVE X IN THEM.
THIS HAS A FACTOR OF X AND THIS HAS A FACTOR OF X.
SO, WE WANT THE TERMS CONTAINING X
ON ONE SIDE OF THE EQUATION
AND THE TERMS NOT CONTAINING X
ON THE OTHER SIDE OF THE EQUATION.
SO, IF WE WANT X TO BE ON THE LEFT SIDE OF THE EQUATION
WE'D HAVE TO SUBTRACT FOR X NATURAL LOG 5
ON BOTH SIDES OF THE EQUATION.
NOTICE IF WE DO THIS, THIS WOULD BE EQUAL TO ZERO.
SO, THE X TERMS WOULD ON THE LEFT SIDE.
SO, LET'S GO AHEAD AND UNDO THIS MINUS 3 NATURAL LOG 3
AT THE SAME TIME
BY ADDING 3 NATURAL LOG 3 TO BOTH SIDES OF THE EQUATION.
AGAIN, NOTICE HOW THIS WOULD BE ZERO AS WELL.
SO, NOW WE'RE LEFT WITH 2X NATURAL LOG 3
MINUS 4X NATURAL LOG 5,
AGAIN, THESE ARE NOT LIKE TERMS, EQUALS.
ON THE RIGHT SIDE,
WE HAVE 5 NATURAL LOG 5 PLUS 3 NATURAL LOG 3.
AND NOW TO SOLVE THIS FOR X,
WE'RE GOING TO FACTOR OUT THE COMMON FACTOR OF X HERE.
SO, WE'D HAVE X TIMES THE QUANTITY 2 NATURAL LOG 3
MINUS 4 NATURAL LOG 5 EQUALS THE RIGHT SIDE.
AND NOW BECAUSE THIS X IS ATTACHED BY MULTIPLICATION,
WE CAN SOLVE FOR X BY DIVIDING BY THIS QUANTITY.
SO, WE'LL DIVIDE BOTH SIDES BY 2 NATURAL LOG 3,
MINUS 4 NATURAL LOG 5.
IF WE WANT TO,
WE CAN PUT THE NUMERATOR IN PARENTHESES AS WELL.
NOTICE ON THE LEFT SIDE, ALL OF THIS SIMPLIFIES TO 1,
SO X IS EQUAL TO THIS QUOTIENT HERE,
WHICH WILL APPROXIMATE ON THE CALCULATOR.
SO, OUR NUMERATOR IS GOING TO BE 5 NATURAL LOG 5,
PLUS 3 NATURAL LOG 3,
SO THERE'S THE NUMERATOR, DIVIDED BY THE DENOMINATOR,
WHICH IS 2 NATURAL LOG 3, MINUS 4 NATURAL LOG 5.
AND IT'S TYPICAL TO ROUND TO FOUR DECIMAL PLACES.
SO, X IS APPROXIMATELY NEGATIVE 2.8749.
THIS WAS QUITE A BIT OF WORK.
SO, LET'S GO AHEAD AND VERIFY OUR ANSWER
BY SUBSTITUTING THE VALUE OF X
BACK INTO THE ORIGINAL EQUATION.
SO, THIS IS TELLING US
THAT IF WE HAVE 3 RAISED TO THE POWER OF 2
TIMES NEGATIVE 2.6749
AND THEN SUBTRACT 3,
THIS SHOULD BE EQUAL TO 5 RAISED TO THE POWER OF 4
TIMES NEGATIVE 2.6749 PLUS 5.
LET'S SEE IF THAT'S ACTUALLY TRUE.
WE'LL FIRST EVALUATE THE LEFT SIDE.
SO, WE HAVE 3 RAISED TO THE POWER OF 2
TIMES NEGATIVE 2.6749 MINUS 3, ALL THIS IS THE EXPONENT
AND NOW, LET'S CHECK THE VALUE OF THE RIGHT SIDE.
AND NOTICE HOW THESE VALUES ARE APPROXIMATELY THE SAME,
BUT NOT EXACTLY THE SAME
BECAUSE WE DID ROUND X TO FOUR DECIMAL PLACES.
BUT THIS SHOULD BE ENOUGH INFORMATION
TO VERIFY THAT OUR SOLUTION IS CORRECT.