Assalamualaikum warahmatullahi wabaraktuh
This video will be discussing on how to
solve equations
involving indices.
There are generally
two types of equations
involving indices.
The first one is
equation
with the same base.
The second one is
equations with different bases.
 
We will discuss
how to solve equations of indices with
different bases
after discussing log in chapter 1.3.
Focusing on equations with the same base
to solve this type of equation,
we're going to use two methods.
The first one is by comparing its power
the second method is by changing it into
quadratic equation.
Let's look at the first example.
We are given 3^(2x-3) = 81.
First thing to do is to recognize its
base.
On the left side here
we have base 3.
On the right side here
we have 81.
We know that 81 is equal to 9^2
and we know that
9 is equal to 3^2
Therefore, we can rewrite the equation as
3^(2x-3)=(3^2)^2
Let's simplify this a bit more.
So we have 
3^(2x-3)=3^4.
Now we know that both left and right
side has the same base
and because we only have one index on the left
and one index on the right
and they both have the same base
we can use the first method that I mentioned earlier.
We can compare its power.
So this power here and this power here.
By comparing,
we can have 2x-3=4.
And suddenly this is very easy to solve.
so we have 2x=7 and
lastly we have x=7/2.
Now this looks scary but don't be.
I'll guide you through it
Remember our first step from the
previous slide?
The first step to know is to recognize
its base
but to recognize the base, you need to
recognize
which ones is the index.
Now this is an
index
and this is also an index
and these two indices
look similar except for this power 2
here.
By law of indices I can rewrite this
as (2^x)^2.
So let me write the question again with this new information.
So (2^x)^2 - 12(2^x) +32 =0.
Now you would think, 
"hm, this does not look any easier."
but, what if we substitute 2^x = a?
then the equation now will be
a^2 - 12a +32 =0.
Now you see that this is just a
quadratic equation
and to solve this quadratic equation,
you need to factorize this to become
(a-8)(a-4)=0
and now we have a-8=0
and a-4=0
and then we have a=8 and a=4.
But we don't want to find ‘a’
we wanted to find the value of ‘x’.
So, we're going to substitute it back.
So 2^x=8
and 2^x=4.
You know that 8=2^3
and you know that 4=2^2.
And again, both sides has the same index
and this one too has the same index
and again by comparing its power,
Therefore,
here x=3
and here x=2.
Now let's test you.
Don't peek.
Pause this video and try these
two questions.
When you're finished, you
can continue this video
and we'll check your answers.
Pause now.
I'll wait ...
Ready? Let's check.
The first thing to do
is to recognize the base.
So we're going to change 49 and 343 into base 7.
Let's rewrite the equation here.
We have 7^(x+2) / (7^2)^x
equals to 7^3.
Now simplify the left side here.
It will become 7^(x+2- 2x) =7^3.
This one becomes 7^(2-x)=7^3.
So by comparing,
we have 2-x=3
then -x =1
and lastly x=-1.
Good job.
Next question.
We can rewrite this as
(3^x)^2=6.(3^x)+27
and if you can recognize this,
this one looks like a quadratic.
Let’s substitute 3^x = a.
So we have a^2=6a+27
We're going to make this equal to 0.
So we are going to move
6a and 27 on the left side.
a^2 -6a -27 =0.
Next we are going to factorize this.
So
(a-9)(a+3)=0.
So, a-9=0 and a+3=0.
Let's look at this solution first a=9.
We don't want to find the value of a.
We want to find the value of x.
so you're going to substitute 3^x back again
so 3^x=9 and we know that
9=3^2.
So by comparing its power,
here we have, therefore, x=3.
(the answer should be x=2, NOT x=3. My apologies)
Now look at this solution here we
have a=-3
Again for the same reason,
we're going to substitute 3^x again
So 3^x=-3.
We're going to reject this solution.
Why? Because the positive base
cannot be equal with the negative base.
So this is the end of chapter 1.1 on
indices.
Thank you for your attention and see you
on the next video.
Have a great day. Take care.
