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- NOW WE'LL EVALUATE TWO LOGARITHMS
WITHOUT THE USE OF A CALCULATOR
WHEN THE NUMBER PART OF THE LOGARITHM IS A FRACTION.
HERE WE HAVE LOG BASE 5 OF 125TH
AND WE HAVE LOG BASED 2 OF 1/16TH.
TO EVALUATE THIS, WE'LL SET IT EQUAL TO A VARIABLE LET'S SAY X.
WRITE THIS AS AN EXPONENTIAL EQUATION AND THEN SOLVE FOR X.
TO WRITE LOG EQUATION AS AN EXPONENTIAL EQUATION,
YOU NEED TO REMEMBER THE PLACEMENT OF THE BASE EXPONENT
AND NUMBER.
B IS THE BASE, "A" IS THE EXPONENT.
AND N IS THE NUMBER.
A NICE WAY TO REMEMBER THIS CONVERSION
IS TO START WITH THE BASE
AND THEN WORK AROUND THE EQUAL SIGN
TO FORM THE EXPONENTIAL EQUATION.
WHAT I MEAN BY THAT IS WE'LL START WITH OUR BASE 5,
5 TO THE POWER OF X MUST EQUAL 125TH.
SO AGAIN 5 TO THE POWER OF X MUST EQUAL 1/25.
NOW WE WANT TO WRITE THE RIGHT
SO THIS EQUATION AS 5 RAISED TO A POWER.
WHAT WE SHOULD RECOGNIZE WE CAN WRITE 25 AS 5 SQUARED
SO WE CAN WRITE THIS AS 5 TO THE X = 1/5 SQUARED
AND NOW USING OUR PROPERTIES OF EXPONENTS,
IF WE MOVE THIS ACROSS THE FRACTION BAR,
OR TAKE THE RECIPROCAL OF THIS,
IT'S GOING TO CHANGE THE SIGN OF THE EXPONENT
SO 1/5 SQUARED IS EQUAL TO 5 TO THE -2.
SO WE'D HAVE 5 TO THE POWER OF X = 5 TO THE POWER OF -2.
NOTICE HOW THESE ARE EQUAL
AND THE BASES ARE THE SAME AND THEREFORE X MUST EQUAL -2.
WELL IF X = -2 THEN WE KNOW LOG BASE 5 OF 125TH ALSO EQUALS -2.
LET'S TAKE A LOOK AT A SECOND EXAMPLE.
WE'LL SET THIS EQUAL TO A VARIABLE LET'S SAY Y.
WRITE IT AS AN EXPONENTIAL EQUATION SO HERE THE BASE IS 2
SO WE HAVE 2 RAISED TO THE POWER OF Y = 1/16TH.
SO 2 TO THE POWER OF Y = 1/16TH.
WE WANT TO WRITE THE RIGHT SIDE OF THIS EQUATION
AS 2 RAISED TO SUM POWER.
LET'S IGNORE THE FRACTION FOR A MOMENT AND TAKE A LOOK AT 16.
16 IS EQUAL TO 4 x 4 AND 4 IS EQUAL TO 2 x 2.
SO 16 IS EQUAL TO 2 TO THE 4TH
SO WE CAN WRITE THIS AS 2 TO THE Y = 1/2 TO THE 4TH
BUT THEN AGAIN USING OUR PROPERTIES OF EXPONENTS
IF WE MOVE THIS UP TO THE NUMERATOR,
OR TAKE THE RECIPROCAL
IT'S GOING TO CHANGE THE SIGN OF THE EXPONENT
SO WE'D HAVE 2 TO THE Y = 2 TO THE -4TH POWER
AND AGAIN THESE ARE EQUAL. THE BASES ARE THE SAME.
THEREFORE Y IS EQUAL TO -4 AND IF Y IS EQUAL TO -4
THEN OUR LOGARITHM OR LOG BASE 2 OF 1/16TH IS EQUAL TO -4.
I HOPE THIS WAS HELPFUL.