- WE WANT TO SOLVE THE GIVEN EXPONENTIAL EQUATIONS.
OUR FIRST EQUATION, WE HAVE 5 RAISED TO THE POWER OF X = 78.
SINCE WE CANNOT WRITE 78 AS 5 RAISED TO ITS OWN POWER,
WE WILL HAVE TO USE LOGARITHMS TO SOLVE THIS EQUATION.
AND BECAUSE WE'LL BE USING THE CALCULATOR,
WE'RE GOING TO USE EITHER A NATURAL LOG OR A COMMON LOG.
IN GENERAL TO SOLVE AN EXPONENTIAL EQUATION,
WHEN THERE'S 1 EXPONENTIAL PART,
WE ISOLATE THE EXPONENTIAL PART
AND THEN TAKE THE LOG OF BOTH SIDES OF THE EQUATION
TO SOLVE FOR X.
NOTICE 5 TO THE POWER OF X
IS ALREADY ISOLATED ON THE LEFT SIDE OF THE EQUATION.
SO FOR THIS EXAMPLE,
WE'LL TAKE THE NATURAL LOG OF BOTH SIDES OF THE EQUATION,
SO WE'D HAVE THE NATIONAL OF 5 RAISED TO THE POWER OF X
= THE NATIONAL OF 78.
REMEMBER OUR GOAL IS TO SOLVE FOR X
SO WE CAN APPLY THE POWER PROPERTY OF LOGARITHMS HERE
AND MOVE THIS EXPONENT OF X TO THE FRONT
SO IT BECOMES THE COEFFICIENT.
SO NOW, WE HAVE X x NATURAL LOG 5 = NATURAL LOG 78.
AND NOW TO SOLVE FOR X,
WE JUST NEED TO DIVIDE BOTH SIDES OF THE EQUATION
BY NATURAL LOG 5.
THIS SIMPLIFIES THE 1,
SO WE HAVE X = THIS QUOTIENT WHICH WE WILL HAVE TO ROUND.
SO NOW, LET'S GO TO THE CALCULATOR,
AND WE'LL JUST TYPE IN NATURAL LOG 78
DIVIDED BY NATURAL LOG 5.
AND IT'S TYPICAL TO ROUND TO 4 DECIMAL PLACES.
IF WE ROUND TO 4 DECIMAL PLACES, THIS WOULD BE 2.7070.
AND WE SHOULD CHECK OUR SOLUTION.
SO IF X IS APPROXIMATELY 2.7070,
THAT MEANS 5 RAISED TO THE POWER OF 2.7070
SHOULD BE APPROXIMATELY EQUAL TO 78.
SO LET'S GO AHEAD AND CHECK THIS.
NOTICE HOW IT IS ALMOST 78.
IT'S A LITTLE BIT LARGER THAN 78, BECAUSE WE DID ROUND UP.
BUT THIS DOES VERIFY OUR SOLUTION.
LET'S TAKE A LOOK AT OUR SECOND EXAMPLE.
HERE WE HAVE 10 RAISED TO THE POWER OF -2X = 2/3.
THE EXPONENTIAL PART IS ALREADY ISOLATED
ON THE LEFT SIDE OF THE EQUATION.
SO NOW, WE'LL TAKE EITHER THE COMMON LOG OR NATURAL LOG
OF BOTH SIDES OF THE EQUATION.
AND BECAUSE HERE WE HAVE BASE 10,
I'M GOING TO TAKE THE COMMON LOG OF BOTH SIDES.
SO WE'LL HAVE THE LOG OF 10
RAISED TO THE POWER OF -2X = LOG 2/3.
AND NOW, WE CAN APPLY THE POWER PROPERTY OF LOGARITHMS
TO MOVE THIS EXPONENT OF -2X TO THE FRONT,
SO WE'D HAVE -2X x LOG 10 = LOG 2/3.
NOW, THE REASON I WANTED TO USE COMMON LOG ON THIS EXAMPLE
WAS BECAUSE THE COMMON LOG OF 10 IS EQUAL TO 1,
SO WE COULD JUST DROP THIS.
BUT I'M GOING TO GO AHEAD AND LEAVE IT IN
JUST IN CASE WE DIDN'T RECOGNIZE THAT.
SO NOW TO SOLVE FOR X, WE'D HAVE TO DIVIDE
BOTH SIDES OF THE EQUATION BY (-2,LOG 10).
NOTICE ON THE LEFT SIDE, -2/-2 SIMPLIFIES TO 1
AS WELL AS LOG 10/LOG 10,
SO X IS GOING TO BE EQUAL TO THIS QUOTIENT HERE
WHICH WE WILL HAVE TO ROUND AGAIN.
SO LET'S GO BACK TO THE CALCULATOR.
AND NOW, WE'RE USING THE COMMON LOG,
SO WE'LL PRESS LOG 2/3 FOR THE NUMERATOR.
2 DIVIDED BY 3 DIVIDED BY, WE NEED AN OPEN PARENTHESIS
FOR THE ENTIRE DENOMINATOR, -2,
OF COURSE LOG 10 IS 1,
BUT I'LL GO AHEAD AND TYPE IT IN.
SO WE HAVE A CLOSED PARENTHESIS FOR THE LOG
AND FOR THE DENOMINATOR, AND ENTER,
SO X IS APPROXIMATELY .0880.
AND TO CHECK THIS,
THIS MEANS THAT 10 RAISED TO THE POWER OF -2 x 0.0880
SHOULD BE APPROXIMATELY EQUAL TO 2/3.
AND 2/3 AS A DECIMAL IS EQUAL TO 0.6 REPEATING.
SO LET'S GO AHEAD AND VERIFY THIS.
AND AS YOU CAN SEE, IT'S VERY CLOSE TO 2/3,
SO OUR ANSWER IS CORRECT.
OKAY, WE'LL TAKE A LOOK AT SOME MORE EXAMPLES
IN THE NEXT VIDEO.