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# logs

– WE WANT TO EVALUATE
THE TWO GIVEN LOGARITHMS WITHOUT THE USE OF A CALCULATOR. WE HAVE LOG BASE 3 OF 81
AND LOG BASE 2 OF 32. TO EVALUATE THESE WE’RE GOING
TO SET THEM EQUAL TO A VARIABLE, LET’S SAY X, THEN WE’LL WRITE THIS
AS AN EXPONENTIAL EQUATION TO DETERMINE THE VALUE OF X. SO TO WRITE THIS LOG EQUATION
AS AN EXPONENTIAL EQUATION WE CAN USE OUR NOTES BELOW
WHERE B IS THE BASE, “A” IS THE EXPONENT,
AND N IS THE NUMBER. ANOTHER NICE WAY TO REMEMBER
THIS IS TO START WITH THE BASE, WORK AROUND THE EQUAL SIGN TO
FORM THE EXPONENTIAL EQUATION. SO HERE WE HAVE 3 RAISED TO
THE POWER OF X MUST EQUAL 81. SO 3 TO THE POWER OF X
MUST EQUAL 81 AND NOW WE’LL SOLVE FOR X. WE CAN DO THIS
WITHOUT THE USE OF A CALCULATOR BECAUSE WE CAN WRITE 81
AS 3 RAISED TO A POWER. 81=9 x 9 AND 9=3 x 3,
SO 81=3 TO THE 4th. SO NOW WE HAVE 3 TO THE X
=3 TO THE 4th. SO THESE TWO ARE EQUAL
AND THE BASES ARE THE SAME, AND THEREFORE THE EXPONENTS
MUST BE EQUAL MEANING X MUST EQUAL 4. WELL IF X=4
THEN LOG BASE 3 OF 81 MUST=4. LET’S TAKE A LOOK
AT A SECOND EXAMPLE. WE’LL SET THIS EQUAL
TO A VARIABLE LET’S SAY Y. WRITE THIS AS AN EXPONENTIAL
EQUATION. SO 2 IS THE BASE, Y IS THE
EXPONENT AND THE NUMBER IS 32. SO 2 TO THE POWER OF Y
MUST EQUAL 32. LET’S TAKE A LOOK AT 32. 32 IS EQUAL TO 2 x 16,
16 IS EQUAL TO 4 x 4, 4 IS EQUAL 2 x 2. SO WE HAVE 1, 2, 3, 4, 5 FACTORS
OF 2 SO 32 IS 2 TO THE 5th. SO 2 TO THE POWER OF Y
EQUALS 2 TO THE 5th. AND AGAIN, THESE ARE EQUAL.
THE BASES ARE THE SAME. SO THE EXPONENTS MUST BE EQUAL
AND THEREFORE Y IS=TO 5. WHICH MEANS OUR LOGARITHM
IS=TO 5. SO WE HAVE LOG BASE 2 OF 32=5. NEXT, WE’LL TAKE A LOOK
AT TWO EXAMPLES WHEN THE NUMBER PART
OF THE LOGARITHM IS A FRACTION.

– WE WANT TO EVALUATE
THE COMMON LOGARITHMS WITHOUT THE USE
OF A CALCULATOR. IN THE FIRST EXAMPLE
WE HAVE COMMON LOG 10,000. REMEMBER WHEN A LOG IS WRITTEN
THIS WAY WITH NO BASE WE KNOW THAT IT’S COMMON LOG
OR LOG BASE 10. AND TO EVALUATE THIS COMMON
LOG, WE’LL SET IT EQUAL TO X, WRITE IT AS AN EXPONENTIAL
EQUATION, AND THEN ONCE WE FIND
THE VALUE OF X WE KNOW THE VALUE
OF THIS LOGARITHM. TO CONVERT A LOG EQUATION
TO AN EXPONENTIAL EQUATION, REMEMBER B IS THE BASE,
A IS THE EXPONENT, AND N IS THE NUMBER. AN EASY WAY TO DO THIS
IS TO START WITH THE BASE, WORK AROUND THE EQUAL SIGN TO
FORM THE EXPONENTIAL EQUATION. SO HERE WE HAVE 10 RAISED TO
THE POWER OF X MUST=10,000. SO 10 TO THE POWER
OF X=10,000. NOW, THE REASON WE CAN DO
THESE WITHOUT OUR CALCULATOR IS BECAUSE WE CAN WRITE 10,000 AS 10 RAISED TO THE SOME POWER
10,000 IS=TO 100 x 100, AND 100 IS=TO 10 x 10, SO 10,000 IS=TO 10
TO THE FOURTH. SO WE HAVE 10 TO THE X=10
TO THE FOURTH. A QUICK EASY WAY TO DETERMINE
THE EXPONENT HERE IS TO COUNT THE NUMBER
OF ZEROS. IF WE HAVE 1
FOLLOWED BY FOUR 0’s THAT’S=TO 10 TO THE FOURTH. SO 10 TO THE X=10 TO THE
FOURTH MEANS THESE ARE EQUAL AND THE BASES ARE THE SAME,
THEREFORE, X MUST=4, AND THEREFORE, LOG 10,000=4.   LET’S LOOK AT AN EXAMPLE NOW WHERE THE NUMBER
IS A FRACTION. AGAIN, WE HAVE COMMON LOGS, WE NEED TO RECOGNIZE
THAT THE BASE IS 10, WE’LL SET THIS EQUAL TO Y. IF WE DETERMINE THE VALUE
OF Y, WE KNOW THE VALUE
OF THIS LOGARITHM. SO NOW I’LL WRITE THIS
AS EXPONENTIAL EQUATION, SO THE BASE IS 10,
THE EXPONENT IS Y, AND IT’S=TO 1/1,000. SO WE HAVE 10 TO THE POWER
OF Y=1/1,000. NOTICE THAT 1,000
HAS THREE 0’s, SO 1,000 IS=TO 10
TO THE THIRD. SO WE HAVE 10 TO THE Y=1/10
TO THE THIRD. AND NOW WE CAN USE
OUR EXPONENT PROPERTIES TO REWRITE THIS. IF WE MOVE THIS
ACROSS THE FRACTION BAR OR MOVE THIS UP
TO THE NUMERATOR, IT’S GOING TO CHANGE THE SIGN
OF THE EXPONENT. SO THIS WOULD GIVE US 10
TO THE Y=10 TO THE -3. AGAIN, IF THESE ARE EQUAL
AND THE BASES ARE THE SAME, THEN THE EXPONENTS
MUST BE EQUAL. THEREFORE, Y IS=TO -3. THEN IF Y IS=TO -3 THEN THE ORIGINAL COMMON
LOGARITHM IS=TO -3. MEANING THE COMMON LOG
OF 1/1,000 IS=TO -3. NEXT WE’LL LOOK
AT TWO EXAMPLES OF EVALUATING COMMON LOGS WHERE WE CANNOT WRITE
THE NUMBER AS 10 TO A POWER. AND THEREFORE, WE’LL HAVE
TO USE THE CALCULATOR. I HOPE YOU FOUND THIS HELPFUL.

– 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.

Hello and welcome to HP Calcs! My name is
Rick and today we will be looking at how to find the log of a number on the Sharp
EL-510 RN calculator. The standard setting for logs on our calculator is log base 10. In order to find the log of a number we will use the second function and 1 buttons. If we want to find the log base 10 of 1000 we would hit second function, 1 To place log on the screen. then enter the number we want to find the log for. In this case 1000. finally we hit the equals button and get the answer 3 which is correct if we want to find a different log base
we would enter it as a fraction. let’s say we want to find the log base 2 of 1,000. We would enter second function log 1000 divided by second function log 2
equals this gives us 9.96578 etc. Which is correct. All right I hope this helps and good luck on those tests!

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– IN THESE TWO EXAMPLES, WE’RE ASKED TO EVALUATE
THE COMMON LOGARITHMS USING A CALCULATOR. THE REASON WE’RE ASKED
TO USE A CALCULATOR IS BECAUSE NOTICE HOW THE
NUMBER PART OF THE LOGARITHMS CANNOT BE WRITTEN AS 10 RAISED
TO AN INTEGER POWER. SO WE’LL HAVE TO GET
DECIMAL APPROXIMATIONS FOR THESE COMMON LOGARITHMS. SO FOR THE COMMON LOG OF 110
WITH THE CALCULATOR WE’LL PRESS THE COMMON LOG
WHICH IS THIS BUTTON HERE, 110, CLOSED PARENTHESIS,
AND PRESS ENTER. WE’RE ASKED TO ROUND
TO FOUR DECIMAL PLACES SO THIS WILL BE APPROXIMATELY
2.0414. BUT I DO WANT TO EMPHASIS
WHAT THIS MEANS. REMEMBER THIS IS
COMMON LOGARITHM, WHICH MEANS
THIS IS LOG BASE 10. SO WHEN EVALUATING A COMMON
LOGARITHM ON THE CALCULATOR WE’RE DETERMINING THE EXPONENT WHICH WE RAISE 10 TO
TO GET THE NUMBER 110. MEANING 10 RAISED
TO THE POWER OF 2.0414 IS APPROXIMATELY EQUAL TO 110. WE KNOW IT’S GOING TO BE
APPROXIMATE BECAUSE WE DID HAVE TO ROUND
THIS VALUE AND SINCE WE DID ROUND UP THIS VALUE SHOULD BE JUST
A LITTLE BIT LARGER THAN 110. LET’S GO AHEAD AND CHECK THIS. WE HAVE 10 RAISED
TO THE POWER OF 2.0414, WE CAN SEE IT’S VERY CLOSE TO
110, THIS A LITTLE BIT LARGER. NOW LET’S TAKE A LOOK
AT OUR SECOND EXAMPLE. WE HAVE THE COMMON LOG
OF 0.425, SO WE’LL PRESS THE LOG KEY
AND 0.425, CLOSE PARENTHESIS, ENTER. THIS VALUE
IS APPROXIMATELY -.3716. AGAIN THIS IS THE ANSWER
THAT THEY’RE LOOKING FOR AND AGAIN I WANT TO EMPHASIS
WHAT THIS MEANS. THIS IS COMMON LOG,
SO IT’S LOG BASE 10 SO 10 RAISED TO THE POWER
OF -0.3716 WILL BE APPROXIMATELY
EQUAL TO 0.425. AND LET’S GO AHEAD
AND CHECK THIS. 10 RAISED TO THE POWER
OF -.3716, AS WE CAN SEE
IS APPROXIMATELY 0.425. SO WHILE IT’S EASY
TO GET THESE VALUES WHAT I THINK IS IMPORTANT
TO REMEMBER WHEN EVALUATING LOGARITHMS WE’RE ACTUALLY
FINDING EXPONENTS. I HOPE THIS WAS HELPFUL.

BAM!!! Mr. Tarrou. In this lesson we are going
to do three examples of graphing logarithmic functions using tables. We are not going to
be allowed to use our calculator. A couple of years ago, two or three years ago I did
a video related to this same topic. I spent a lot of time explaining the inverse relationship
between exponential functions and logarithms functions. I will have a link to that lesson
in the description. I finished off with an example very similar to to how we are going
to work these problems. But then the last one or two examples I did was just sketching
these functions off of transformations. You know, knowing the parent function and moving
it left and right, up and down, flipping it, and whatever. Well if you don’t know what
the parent function looks like, maybe you know what y equals log base two looks like,
but what happens if the base is equal to one half? What happens if there is a leading coefficient
or a plus at the end? Or maybe there are so many transformations involved in the equation
that your sketch is uh…of these functions just based on transformations and knowing
what the log, or excuse me…the parent function looks like, just ends up being a little bit
rough. I don’t want a rough sketch, I want a really nice graph. At least know five points
that these go through and find those five points without the aid of a calculator and
again get that really nice accurate graph, not a rough sketch just based off of transformations.
As we finish these three examples we will talk about the domain and range of these graphs
as well. We are going to put these first two graphs on the same x y coordinate system because
I just want to point out the differences and effect of when we have a base that is larger
than one and what your logarithmic graphs look like when your logarithm has a base that
is between 0 and 1. We also have a vertical stretch which is an a value which is equal
to three. But, I really want to talk about what happens with these different bases. Remember
that you cannot take the log of a negative number. I am just reviewing some logarithmic
facts for you. The answer that you get from a logarithm is an exponent. Hopefully if you
are watching this video you already know that. I like to mention that the common log, log
base ten of a hundred is equal to two because the base of that log is 10, common log, ten
squared is a hundred. I am pointing in here like it is base ten. If I do log base ten
of 100, we get an answer of 2. That is an exponent. It is ten squared gets you back
to a hundred. There is at least a five second review of how logarithms are related to exponents.
Ok, so when you look at logarithms…when you convert from log form to exponential form
you need to remember of course that the base of the logarithm is going to be the base in
exponential form. You do get exponents from logarithms. So we will have say two to the
y, and then the x plus one is an answer, not as far as far as having this written in logarithmic
form but it will be the answer when you get this written in exponential form. That is
exactly how we are going to graph these three examples without the aid of a calculator.
We are going to convert them into exponential form, build up a t-table, pick out some smart
numbers so that we can do it in our head, and plot those points and get our graph. Now,
just one more time. If you do know your parent functions and you know your transformations,
then you will know the difference between y equals log base 2 of x plus one and y equals
log base two of just x. It is a horizontal shift to the left or a horizontal translation
to the left of one unit. So let’s see what this looks like though doing this problem
with a t-table. We have a base, we have an exponent, and we have an answer. That is again
how we are going to do these problems without the aid of a calculator. So we have two to
the y power is equal to x plus one. We see that y variable is up here in the exponent
and is it is sort of where the math is happening. We are going to therefore… We are not going
to re-isolate the variable y, it already was in the first place. I guess you could make
it a log base ten or log base e, but again I want to do this without the aid of a calculator.
I can take the base of two and raise it to certain powers in my head. So we are going
to solve for x, and we are going to do that by subtracting both sides by one. AND we get
two to the y minus one is equal to x. Now I said that we are going to do this with tables.
Normally when you do a t-table you pick the x values, but that is normally the y is along
and the x is over there where the math is happening. Well now the x is alone and it
is the y that is involved with all the math. Two to the y power, now there is a minus one
with it. So when we build up our t-table we are going to be picking values of y. Now what
values do you want to use? Pick some small numbers around the zero range and use integers
more than likely. We want to pick some numbers that can easily raise two to that power. Let’s
just say that y is going to be negative two, negative one, zero, one, and two. I don’t
usually just use two or three points for graphing a function that is curving. Right, what direction
is it curving, how sharply is it doing that? So let’s make up some…let’s do some work
here. So y is equal to negative two. That means that we are going to have two, the base
of two to the exponent of negative two minus one is equal to x. Two to the negative two
power. How do you deal with, or get rid of a negative exponent? How do you rewrite this,
or move this base around so this exponent is positive two as opposed to negative two?
Well you move the base. If the base is in the numerator with the negative exponent,
you drop it down. If the base is in the denominator with a negative exponent, you will move it
up to the top of the fraction. And if you understand division and the properties of
exponents, you can hopefully understand how moving the base up and down will change the
sign of the exponent. But, this will be one over two squared minus one. That is going
to be two squared of course is four. Now we are looking at 1/4 minus one. When you start
dealing with fractions you should make everything look like a fraction. So now we need a common
denominator. We are going to multiply the numerator and denominator of the one by four.
We have now one over four minus four over four is going to be equal to negative three
fourths. When we have a y value of negative two our x value is -3/4. When we let our y
value be negative one, we are going to have, for this next line, we are going to have two
to the negative one. We are going to have to rewrite this to get rid of the negative
exponent, as one over 2 to the first, or just one over two minus one over one. Now we are
going to combine these fractions. Multiply the numerator and denominator by two giving
us 1/2 minus 2/2. When you have common denominators, you just add or subtract those numerators.
One minus two is equal to negative one….half. Now we are going to let y equal zero. And
anything to the zero power is going to be equal to one. We are looking at one minus
one which is going to be equal to zero. Then we are going to let y equal one. And this
is going to go a lot quicker because we are not dealing with fractions any more. Two to
the first is two minus one is one. And then when we let y equal two… I guess I could
have rewritten that line as I was going… Two squared is four and four minus one is
three. That was pretty easy. I don’t need a…require a calculator to do that. And…
I can’t do base of two!!!:( My calculator only does base e and base ten. Who cares.
You don’t even need a calculator. So let’s see here. X is going to be -3/4. Each of the
tick marks count for one. X is equal to -3/4 and y is going to be -2, so I am going to
try my best to estimate that. We have an x values of -1/2 and a y value of -1. We are
passing through the point of (0,0). Logarithmic functions, if there has not been a horizontal
or vertical transformation, we have a plus or minus here inside the math function to
move the graph to the left, if there has been no horizontal or vertical translation or slide,
log functions will always pass through the point of (1,0). Then we have (1,1) and (3,2).
Now, by definition logarithms must have a base which is positive. If they had a base
which was negative, it is nice to talk about it now that we have this in exponential form.
A negative number raised to a power, again we are talking about functions right…some
kind of graph that would look like a function and pass a vertical line test… If you have
a base which is negative and you raise it to an even power, the answer is going to be
positive. Forget the plus or minus something constant going on. Then when you take that
negative base, that base of the logarithm is the base of your exponent, a negative number
raised to a…what did I just say? An even number is positive and an odd power is negative.
So if you had that negative base, or a negative base in an exponential function the graph
would be oscillating back and forth violently between positive and negative values and you
wouldn’t get a smooth curve or function. So that is why your bases have to be positive.
If the base of your logarithm has to be positive and what you get out of it is an exponent,
so two to something has to be equal to what is inside that parenthesis, you are not going
to take a positive number raise it to any power… I just took, I erased it, but I took
a positive base of two and raised it to the negative two power, I didn’t get a negative
answer. That is negative but that was after I subtracted by one. You are never going to
take a positive base raise it to any power and get a negative answer. So you cannot log
zero and you cannot log a negative number. So logarithms are always going to have a vertical
asymptote as part of their graph. We can see that the graph is curving down and the first
one which I really intended on doing in yellow is going to look like this. Instead of the
vertical asymptote being at x equals zero it is going to be at x is equal to negative
one because of that horizontal shift to the left. I didn’t even have to think about that!
The table of values allowed us to see that developing and our graph looks something like
this. Now, it is sort of a sketch and sort of a graph. I mean in this part of the graph
it is really accurate but now…you know..I am not sure how quickly it is going to climb
as we go to the right. But it is going to look something like that. So now we have got
our first graph done. Y is equal to log base two of x plus one. Now, you might have thought,
“I know what the parent function looked like. I knew it was going to move to the left anyway.”
But, that is not just a sketch, it is more of an accurate graph. Do you know the parent
function for a base of one-half, and with that vertical stretch of three. So, let’s
do this with a t-table. How is that base of 1/2, the base being between 0 and 1, going
to compare to when we had a base which is bigger than one. Let’s just put into exponential
form. We are going to… You cannot put this into exponential form until you get the log
function alone. I want to have a leading coefficient of one, and then convert it into exponential
form. I could bring the three up as a power and say that it is y equal log base 1/2 x
to the third and then convert it into exponential form. But my y is going to have some kind
of exponent and my x is going to have some kind of exponent, and I want to just plug
some values in for y and get x. You know, like get the answer for x. I just want the
x alone. So we are going to divide both sides by 3. I don’t know why I picked up green when
the problem is in purple. But we have y over three is equal to log base one half x. Now
we have x is nice and alone. That is going to be handy for when we make our t-table.
The leading coefficient of the logarithm is equal to one. So now we are going into exponential
form which is base, exponent, answer. So we have one half raised to the y over three power
is equal to x. Now you think, well that looks a little bit nasty but you have control when
you are building a t-table of what you are decide to set for your values of y. Your y
is your independent variable and x is dependent variable, at least the way the equation is
structured. So when we set up or t-table, and I want to do this without the aid of a
calculator why not just pick values of y that you can divide by three like 6, 3, and 0.
So we have x and y and y is going to be negative six, negative three, zero, three, and six.
Now let’s plug these in and find out what happens. Let’s see. Let’s get the negatives
out of the way first. We are going to have negative six, so negative six divided by three
is going to be a power of one half. That is going to be one half to the negative two.
If you want to rewrite your base so that it does not have a negative exponent you flip
that base. We are going to have two to the positive two power. Two squared is equal to
four. Repeat the process with negative three and we are going to have negative three. Negative
three divided by three, that is coming in from here right, is equal to negative one
and that is going to be equal to two over one to the first which is of course just equal
to two. Now, when we put in that y value of zero, we are going to have… Letting y equal
zero we have 1/2 to the zero over three power. Zero can be on top, it just cannot be in the
denominator. Zero divided by three is zero. We have one half to the zero power which is
anything to the zero power is one. That is true because…let’s see here. How about,
what is x to the third…this is just an example of…what is x the third divided by x to the
third? We know that anything divided by itself is equal to one right? Well when you divide
like bases and if I didn’t want to just say …ohh anything divided by itself is one…When
you divide like bases, don’t you subtract the exponents? Wouldn’t that be x to the zero
power because three minus three is zero? Yeah anything to the zero power is is one. I don’t
if you knew that or not. Some of my teachers in high school used to just always say, “It
is just a rule, memorize it.” We have, opps I need to put the answer there. Now we are
going to let y equal three. Look at that, I already have the bottom of the three in
there. We are going to let y equal three. Three divided by three is equal to one and
we have one half to the one power which of course is one half. When we let y equal six
we are going to have six divided by three which is two. One half squared is one half
times one half. I don’t think you really need to see that expanded, but in case you do one
times one is one and two times two is four. Alright, so let’s see here. When x is equal
to, we are going to do this in purple for the graph, when x is equal to one y is zero.
I don’t know why I am doing that point first. When x is equal to 1/2 y is equal to three.
So when x is 1/2 y is equal to three. See how the graph is not rising up to the right?
It seems to be falling. When x is equal to 1/4 the y coordinate is all the way up to
6. When x is two, the y value is negative three. And when our x value is four the y
coordinate is negative six. So x is four, and y is negative six. That graph is going
to look something like. More than something like this, it is actually going through these
points, but I don’t know how quickly it bends over as it goes to the right. To answer that
I would need to make a bigger t-table. But, hmm… Ooo, I said I was going to talk about
domain and range. Now notice that the base being between zero and one, this graph is
falling. Let me check the time by the way. Ok, so with that base bigger than one, and
most of the time your bases are going to be bigger than one, those logarithms come up
from negative infinity and go over to the right. With that base of one half, that is
going to get a function that is monotonic decreasing as opposed to monotonic increasing.
It is interesting. Now the domain and range of this yellow graph. The domain is going
to be, well it goes all the way to the left and it approaches negative one, but x equals
negative one is a vertical asymptote. The x’s that make up this graph, this yellow graph,
are never going to become negative one but it will approach negative one. So the domain
are your x’s between negative one, oops almost wrote infinity, negative one to infinity.
The graph as you can see goes down and it is sloping over sharply to the right but it
will never stop increasing. So this yellow graph is going to go up and down forever,
so the range which are the y values are going to go from negative infinity to positive infinity…
AND BEYOND! The purple graph, well it goes up and down forever as well so it’s range
is going to be the same. It’s domain though, we have a vertical asymptote at x equals zero
so this purple graph is going to have a domain of zero to infinity. Now I just have one more
example and really it is same kind of deal, just a more complicated example. So I am going
to step off, put that question up here and just reveal the solution one step at a time
so that you could pause the video, try it on your own, and see if you can’t get that
logarithmic equation written into exponential form and check your work. Because I am going
to just reveal the next example in small steps at a time, because really the process is the
same. Be right back, BAM!!! Nanananana… So we have our last example. It was y is equal
to negative two plus log base three of x plus four. As you saw, trying to isolate that logarithm
we added both sides by two. Then we went into exponential form, base-exponent-answer. And
then we had to subtract both sides by four until we completely isolated the x. Ok, I
cheated a little bit and used a calculator to figure out that 35 over 9 was 3.9 so I
did not have to do the long division by hand and so on for the remaining decimal. I did
really intentionally want my x values to be so close together and be at least a tiny bit
difficult to graph. I just looked at the exponent of y plus two and thought, well why don’t
I pick my y values so that the middle term is going to have an exponent of zero, and
we will have a couple of negative exponents and a couple of positive exponents. That is
how I came up with the idea of using these y values for my t-table. Also, i went ahead
and before putting the vertical asymptote on our graph, what is inside the logarithm
again has to be positive. It has to be greater than zero…greater than and not equal to.
x plus four is greater than zero, subtract both sides by four and we get the placement
of our vertical asymptote, the domain, put the points down, and that is the end of our
last example. I hope this helps. I am Mr. Tarrou. BAM!!! Go do your homework:D AND HAPPY
PI DAY!!!