Bam!!! Mr. Tarrou. I recently came to the realization that all of my videos, my lessons
that help you understand or show you how to solve equations with logarithms, I never included
an example where those logarithms had a different bases. So here we go. We are going to solve
log base 4 of x+2 minus log base 16 of x^2-4 is equal to one. Now, when you have equations....
You know, I have examples in my other lessons which I will put links to in the description
of this one, where we have multiple log functions in the same equation but they always had the
same base. So that is what we are going to have to do with this one. We are going to
have to rewrite these logarithms to have that same base. Then we will be able to compress
those expressions... or condense those logarithms so there is only one log function. Then we
can rewrite that equation into exponential form and finish the problem. So we see that
we have a log base 4 and a log base 16. So we could write that 16 as 4 squared and see
that our bases are almost the same. But then what do you do with that exponent of 2 that
is on the 4 in the base of the logarithm. So that... If you know there is probably some
kind of nice easy trick to allow you handle that exponent. But if you are not doing logarithms
all the time, you probably don't know any easy tips or tricks to kind of... you know...
jump right to the answer. So what you want to do is use the change of base formula. Now
the change of base formula says that... ah... that I can rewrite a logarithm into any base
log that I would like. I am just going to take the log of whatever base I would like
of the x^2-4 and then divide that by the log base of whatever I want to go into of 16.
So we are going to rewrite this with a base of 4. We will have log base 4 of x+2 minus
the log base 4 of x squared minus 4 over the log base 4 of 16. That is going to be equal
to one. If you would like a review of the change of base formula and not me just sort
of telling you what to do I will put a link to that lesson in the description of the video
as well. Well now let's remember that when you take the log of a number you get an exponent.
So when you take the log base 4 of 16, that answer is going to be an exponent. The exponent
is going to be applied or be able to be applied to that base and be equal to the value that
you are trying to take the log of. So the log base four of sixteen is equal to two.
But if you are not sure of that. If you are still, even though we are doing some rather
sophisticated problems here, a little bit uneasy on logarithms you can write this in
exponential form and go base... exponent... and at least in exponential form the answer.
We have 4 to the y power is equal to 16. That is 4 to the y power is equal to 4 squared.
There is your y value of 2. So this denominator is equal to 2. I am also going to pull that
out front and we have the log base 4 of x+2 minus 1/2... again that denominator of 2...
that division of 2 coming from the answer of log base 4 of 16... times the log base
4 of x^2-4 is equal to 1. Ok. Well now I have still two log functions in my equation but
they have the same base. That is going to allow us to condense those logarithms. But
I can't do that until I deal with my coefficient of one-half. The power property... the power
rule is going to allow me to bring that 1/2 up on to what I am taking the log base 4 of.
So now we have the log base 4 of x+2 minus the log base 4 of x^2-4 to the 1/2 power.
That is still equal to of course 1. Now with your fractional or rational exponents we know
that the top number is the power and the bottom number is roots. So the power of 1/2 is equivalent
to... or can be rewritten in radical form as the square root. So we are going to condense
these log functions into only one, and I am going to go ahead and write that as the square
root of x^2-4. So we have with those like bases subtraction. You know, when do you again...
Ok, logarithms give you exponents. When do you subtract exponents? When you are dividing
like bases. So that is going to be the log base 4 of x+2 over the square root of x^2-4
is equal to 1. So we are getting close to getting to our answer. Common mistake. x squared
is a perfect square. 4 is a perfect square. So a lot of students will want to square root
each of those individual terms. If they were factors, if they were touching then it would
be just 2x. But with them being two terms inside that radical you cannot take the square
root of each of those individual terms separately of course. So now that we have our equation
with only one log function we are going to make both sides of the equation an exponent
of 4, or just simply put it into exponential form. So we have 4 to the first power is equal
to x+2 over the square root of x^2-4. You know, I kind make it look like... I almost
didn't write that exponent of 1, but for teaching purposes for the first. Of course that is
equal to 4. We are going to now look at... well we are trying to solve an equation that
has a radical in it. Ok. When you are tying to solve an equation with a radical in it,
you want to isolate that radical, um, actually I can deal with it in this format. But I am
going to go ahead and multiply both sides by the square root of x^2-4 just because you
can't solve for a variable when it is in the denominator. It has to come out of there eventually
anyway. I can divide both sides by 4 and eventually I am going to probably need to but for right
now I am going to go ahead and square both sides of the equation. I have just got that
one single term with the radical in it... the square root symbol in it. So we are going
to do the inverse math operation of square root which is square both sides. It is going
to give us x squared plus 4x plus 4 is equal to 16 times x squared minus 4. Ok. We have
parenthesis in the right hand side of our equation. So we are going to distribute the
16 and get x^2+4x+4 is equal to 16x^2... now we have 4 times 10 is 40 and 4 times 6 is
24. 40 and 24 is 64. We now have an equation which is a quadratic. We have powers of 2
and powers of 1 and we have constants. So of course to solve a quadratic to need to
get that thing set equal to zero. So we are going to move the x^2 over with subtraction.
We are going to subtract both sides by 4x. And finally we are going to subtract both
sides by 4 and get negative 68. That is equal to zero. Just to make sure that I am not making
any careless mistakes here. I am doing this math off the top of my head. 15x^2-4x-68.
Excellent. Now, I am not going to attempt to factor this equation, certainly now with
a leading coefficient and a constant that is this large. We are going to go ahead and
identify a, b, and c and plug these values into the quadratic formula. So we have x is
equal to the opposite of b, and the opposite of negative four is going to be positive four.
But I will go ahead and write that with all of the signs in there. That plus or minus
the square root of b squared minus 4ac all over 2a. Oops... 2a.. haha A is equal to 15.
Opposite of b plus or minus the square root of b squared minus 4ac all over 2a. Now simplifying
that we have 4 plus or minus... right off the top of my head -4 squared is 16 minus
4 times 15 times -68... that is equal to the square root of 4096. Yeah right, off the top
of my head! It is on the side chalk board there. 4096. All of that is divided by of
course 30. Sorry about running out of room here. I have my Christmas lights up. The square
root of 4096 is equal to 64 and we are looking at 4 plus the square root of 4096 again like
I just said which is 64 over 3 and we have (4-64)/3. This first fraction is going to
come out to be 68 over 3 and that I don't believe reduces. That would be thirty! That will reduce. That
will be 34 over 15. I apologize for that. Over here we have negative sixty over thirty
which of course is negative two. One thing to be aware of or be careful of, when you
are solving equations with natural... or log functions, or even radicals like square root
and fourth root, probably some trig functions in there. There is plenty of examples of functions
that have a restricted domain. So we have a positive answer and a negative answer and
our original problem has logarithms in it. Well I cannot take, you cannot take the log
of a negative number. Of course again when you log a number you get an exponent. It is
the exponent of the base that will get that... that exponential expression to equal whatever
it is you are taking the log of. And you are not going to raise a positive base, because
the base of logarithms have to be positive so you get a continuous curve... function,
you are never going to raise a positive number to any power positive or negative and get
a negative answer. So you cannot log a negative number. Well, when you plug in... You also
cannot log zero. When you plug -2 in here, -2 plus 2 or -2 squared is positive 4 minus
4 is zero... you cannot log zero either. So one of our solutions is extraneous and 34/15
is our solution to this problem. That is how you solve log equations that have different
bases within the same function. I am Mr. Tarrou. BAM! Go do your homework:D