- 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.
