– 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

AS WELL AS THE GRAPHS HELP YOU BETTER UNDERSTAND

THE DOMAIN AS WELL AS THE EQUATION

OF THE VERTICAL ASYMPTOTES.

## 10 Comments

## Jamie Waldorf

Thanks!! 🙂

## Karamat Qadri

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!

## Bee Radford

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

## Kyle Taran

Wonderful explanation! Very helpful indeed.