 – WE WANT TO FIND THE EQUATION
OF THE VERTICAL ASYMPTOTE TO EACH LOG FUNCTION. OUR FIRST FUNCTION IS F OF X
=LOG OF THE QUANTITY X – 4. NOTICE HOW THERE’S
NO BASE GIVEN. SO WE KNOW THIS IS BASE 10
OR A COMMON LOG. SO WE COULD REWRITE THIS
AS Y=LOG BASE 10 OF THE QUANTITY X – 4. WE SHOULD KNOW BY NOW
THAT THIS IS EQUIVALENT TO THE EXPONENTIAL EQUATION 10 RAISED TO THE POWER OF Y
=THE QUANTITY X -4.   SO WE KNOW X CAN BE ANY VALUE
THAT WOULD BE EQUAL TO 10 RAISED TO THE POWER OF Y, BUT 10 RAISED TO THE POWER
OF Y IS ALWAYS GOING TO BE
POSITIVE. THAT’S THE REASON
WHY TO FIND THE DOMAIN, WE SOLVE THE INEQUALITY
X – 4>0,   AND THEN TO FIND THE EQUATION
OF THE VERTICAL ASYMPTOTE, WE NEED TO SOLVE THE EQUATION
X -4=0. SO IF WE SOLVE THIS INEQUALITY
FOR X, WE WOULD ADD 4 TO BOTH SIDES. THE DOMAIN IS X>4, AND THEN IF WE SOLVE
THIS EQUATION FOR 4, WE’D HAVE X=4, WHICH WOULD BE THE EQUATION
OF THE VERTICAL ASYMPTOTE. SO NOTICE THAT THE X VALUE
CAN GET CLOSER AND CLOSER TO POSITIVE 4. IT JUST CAN’T=+4, BECAUSE IF IT DID, WE’D HAVE
10 TO THE POWER OF Y=0, WHICH CANNOT OCCUR. NOW, WE’LL TAKE A LOOK AT THIS
GRAPHICALLY AS WELL IN JUST A MOMENT. BUT LET’S TAKE A LOOK AT OUR
2nd EXPONENTIAL FUNCTION. WE HAVE F OF X=NATURAL LOG
OF THE QUANTITY X + 1. REMEMBER, NATURAL LOG
IS LOG BASE E. SO WE COULD WRITE THIS
AS Y=LOG BASE E OF THE QUANTITY X + 1, AND THEN IN EXPONENTIAL FORM, WE’D HAVE E RAISED
TO THE POWER OF Y=THE QUANTITY X + 1. SO IN THIS FORM IT’S EASIER
TO SEE THE DOMAIN WOULD HAVE TO OCCUR
WHEN X + 1>0, AND THEREFORE, THE EQUATION
OF THE VERTICAL ASYMPTOTE WOULD OCCUR WHERE X + 1=0. AGAIN, FOR THE DOMAIN WE SOLVED THE INEQUALITY
X + 1>0, AND THEN FOR THE VERTICAL
ASYMPTOTE WE SOLVED THE EQUATION
X + 1=0. SO SUBTRACT 1 ON BOTH SIDES. WE HAVE X>-1 FOR THE DOMAIN, AND THEN WE SUBTRACT 1
ON BOTH SIDES OF THE EQUATION. WE HAVE X=-1
FOR THE EQUATION OF THE VERTICAL ASYMPTOTE. LET’S VERIFY THIS GRAPHICALLY
AS WELL. AGAIN, OUR FIRST LOG FUNCTION
WAS COMMON LOG OR LOG BASE 10. SO IN EXPONENTIAL FORM
WE WOULD HAVE 10 TO THE POWER OF Y
=QUANTITY X – 4. TO GRAPH THIS BY HAND,
WE’D SELECT VALUES OF Y AND THEN SOLVE FOR X, WHICH WOULD PRODUCE
THIS BLUE GRAPH HERE. NOTICE HOW THE GRAPH
APPROACHES +4 BUT NEVER REACHES IT, AND THAT’S WHY WE HAVE A
VERTICAL ASYMPTOTE OF X=+4. JUST KEEP IN MIND
THE GRAPH NEVER DOES TOUCH THIS VERTICAL LINE, X=4, AND THAT’S THE REASON
WHY OUR DOMAIN IS X>+4. AND THEN FOR OUR
2nd LOG FUNCTION, THE NATURAL LOG FUNCTION,
IN EXPONENTIAL FORM WE WOULD HAVE E RAISED
TO THE POWER OF Y=THE QUANTITY X + 1. WE TAKE A LOOK AT THE GRAPH
OF THIS FUNCTION. HERE’S OUR VERTICAL ASYMPTOTE
OF X=-1. THE GRAPH APPROACHES THIS
LINE BUT NEVER TOUCHES IT, AND OUR DOMAIN IS X>-1. OKAY, HOPE THE EXPLANATIONS
THE DOMAIN AS WELL AS THE EQUATION
OF THE VERTICAL ASYMPTOTES.

• ### Jamie Waldorf

Thanks!! 🙂

• haha you teach way better than my math teacher, she sucks! This is way easier than she made it seem.

• ### Sannin Legend

Helped a lot!

• Thank you!!

• ### c00lbandit1

Much thanks

• ### JaeCoxx

People like you make this world a better place!

• ### Vipstephen

This was a LIFE SAVER. Thank you so much for making this video!! You've relieved a lot of stress.

• ### Michael Bruce

WOW, after an hour of manually testing x and y values and self teaching this I stumbled upon this video which completely solidified the concept. Thank you kindly! I subscribed and liked!

• ### Alex Qiu

Thanks

• 