Welcome, ladies and gentlemen.
So what I'd like to do is show you
how to solve logarithmic equations by converting
to exponential form.
And the main basic thing of this is
remember when we worked on converting from exponential
to logarithmic form?
Yes, it was very important for understanding
what logarithms were, and how they're related to exponential.
But for solving logarithmic equations,
it's essential for us to understand
how to convert them back into there,
because here, I have a variable log base 3 of x equals 4.
I need to solve for x.
Well, I can't just type in that into my calculator
and solve for the x.
I got to be able to somehow find a way to isolate the variables.
So I have to get x.
You notice that x is inside of my function.
Its log base 3 of x.
So I got to get x outside of the function.
Well, by converting them rewriting
my logarithmic equation in exponential form,
I can rewrite this as 3 to the fourth power equals x.
So now, to solve for x, all I simply need to do
is take 3 to the fourth power.
Well, you could plug in into your calculator,
or you could also have been working on this for a while
and understand that 3 to the fourth power is 81.
So x equals 81.
OK, the next one again trips up a lot of students
because they don't see the base, and they're like,
what is the base for ln?
Remember the base for ln is e.
So again, I can still do the same thing.
I'm just going to rewrite this in exponential form.
So it's e to the third power equals x.
However, I can't take e because e is an irrational number.
So I am going to use my calculator for this one.
So in my calculator, I'll just do
second natural log, which will give me e raised to a power.
And I'll just type in 3.
And I get approximately-- x is approximately-- I'll round this
to the nearest tenth.
20.1.
OK, so now we have some expressions.
And that's what we're going to get into.
Now are the main important thing I
want to drill home with you on these types of problems
is when we're solving a logarithmic equation
by converting it to exponential form,
we have to have the logarithm isolated, right?
The logarithm cannot be adding or subtracting by the number,
or multiplying or dividing by any number.
In this case, you can see it might
look like you're adding 25.
But that's not the case.
You're actually adding 25 inside the function.
See, that's inside the parentheses.
So that's OK.
Now all I need to do is just convert it to exponential form
because that's inside the function.
It's not outside the function.
The function is isolated.
So by rewriting this, I have 2 to the fourth equals x plus 25.
Now when I take 2 to the fourth power,
I get 16 equals x plus 25.
And then I just solve for x.
So I subtract 25 on both sides, and I get negative 9 equals x.
X equals negative 9.
Now I should actually state that we should always
check for extraneous solutions.
And what I mean by that is if you
look at the graph of a logarithm,
notice that the x values are negative numbers.
So we can't take the logarithm of a negative number.
And a lot of times, this gets scary for people
when they see their solution is negative.
However, if I plug negative 9 back in for x, negative 9
plus 25 is 16.
So I'm still taking the logarithm of a positive number.
When I plug 20 in for x, I'm still
taking the logarithm of a positive number.
Plug 81 in for x, I'm still taking the logarithm
of a positive number.
So as long as I'm not taking the logarithm of a negative number,
we're good.
There's no extraneous solutions.
And we'll get into that more next video.
All right.
So in this case, you see I'm multiplying
by 2 inside the function, but I'm also multiplying by 6
outside the function.
So my function is not isolated.
So to isolate it, what I need to do
is divide by 6 on both sides.
I don't need to divide by 2 because that's
inside the function.
So I have ln of 2x is equal to 5.
Now I'll just do the exact same thing.
Convert it to exponential form, then divide by 2 on both sides.
So I have x is equal to e to the fifth divided by 2.
So I do e raised to the fifth divided by 2.
And I get approximately 74.2.
OK, so in the next example, now you
can see my logarithm is being added by 7
and is being multiplied by 3.
So remember, when using your inverse operations,
you always want to make sure that you undo
addition and subtraction first.
And then undo a multiplication and division.
So therefore, I get log base 2 of x is equal to negative 1/3.
Now to solve for x, now my logarithm has been isolated.
I'm just going to do 2 to the negative 1/3 is equal to x.
Please remember your rules of negative powers,
that one raised to the x to the m
is equal to x to the negative m.
Then we don't need the 1 one property here.
But let's do another rule of exponents.
Remember that x are rational powers.
A to the b is equal to the b root of x to the a.
So in reality, this is 1 over 2 to the 1/3
equals x, which is 1 over the cube root of 2.
Now to do my little calculus, we could
rationalize the denominator.
But we're looking to solving it.
So I'm actually just going to plug this in.
So I'm going to type in 1 divided by the cube root.
Actually, I think I have the cube root in there.
I do.
The cube root of 2, and I get 0.79.
And that's going to be irrational.
Actually, just rounding it to the nearest 10th
is going to be 0.08.
So you could rationalize the denominator
if they're looking for exact answer.
But I'm not looking for exact answers.
We're looking for approximating the solutions.
Now on this problem, I like this one
because again, I am taking the square root,
but the square root's inside the function.
So I don't have to do anything with it.
I can just leave it as it is.
Now you could rewrite this as rewrite this
as a fractional power.
So I put the fractional power in front and then get rid of it.
But there's really no need to do that.
We just have e, right?
So we rewrite this in exponential form.
E to the first equals the square root of x plus 3.
Well, I am still going to want to rewrite this.
To undo the square root I can square both sides.
Actually, let's do that subsequent list.
So e to the first power equals square root of x plus 3.
So to get rid of the square root, I just square both sides.
So therefore, I have e squared equals x plus 3.
And then I'll just subtract 3.
So x equals e squared minus 3.
So then what I'll do is just take my calculator
and just do e squared, and then minus 3.
4.38.
X equals 4.4, as I round.
And I forgot, ladies and gentlemen.
Let's double check.
When I plug in 74 in for that, that gives me positive.
When I plug in 0.8 in for x, that gives me positive.
When I plug in 4.4, 4.4 plus 3.
Square root, OK.
We can still do that.
In the next example, let's just do these inverse operations
quickly.
Subtract three on both sides.
Negative log.
Just remember there's that negative there.
That's why I chose this problem, because a lot of students
will get stuck with this.
They'll forget there's a negative.
So divide by negative 1, divide by a negative 1.
Remember, there's no base there.
So therefore, we know that the base is going to be base 10.
So I'll just write it in there to remind you.
OK, now by rewriting this in exponential form,
I have 10 to the negative second equals x plus 2, minus 2,
minus 2.
OK, so 10 to the negative second is the same thing
as 1/10 squared minus 2.
So I'm just going to do 1 divided
by 10 squared is 100, and then minus 2,
which gives me negative 1.99.
So x is approximately negative 1.99.
Now I'm not going to round this one, at least to the nearest
tenth, because if you notice, if I have negative 1.99,
if I plug that in for x, that's just barely going
to give me enough.
But it still is going to be positive.
So therefore, that is a solution to this case.
So that works.
10 to the negative second minus 2.
Cool.
In our last example-- sorry about that.
Let's see.
What did I-- OK, this is in here.
OK.
So this example, what we're going to do
is, again, just use our inverse operations.
So I'm going to multiply by 3 over 1,
which is the reciprocal here.
So therefore, I have log base 5 of 12x plus 1 equals 6.
Rewrite this in exponential form.
I have 5 to 6 power equals 12x plus 1.
So 5 raised to the sixth power.
5 raised to the sixth power.
OK.
So then, I'll just subtract 1.
So 5 raised to the sixth power.
OK, and then equals 12x plus 1.
Subtract 1, subtract 1.
15624 equals 12x.
And then I'll just divide by 12.
And therefore, I get x equals 1302.1, approximately.
And then again, if I was to plug that back in there,
I would be all set.
So there you go, ladies and gentlemen,
that is how you solve logarithmic equations
by rewriting them in exponential form.
Thanks.