Welcome, ladies and gentlemen.
So what I'd like
to do is show you
how to solve exponential
equations now using
your calculator.
And the reason why we're going
to be using our calculator
is because we can't use
the one to one property.
If you remember, the one to
one property said if a to the x
equals a to the y,
then x equals y.
And the [? Moore's ?] formula,
our basic example I always went
through was if you had
something like this-- 3 squared
equals 3 to the x--
then what does x equal?
Well, then we can
always say x equals 2.
And the reason why it
works is because whenever
your bases are the same,
your powers are equivalent.
And what we did previously
is we had something like 4
to the x equals 16.
So therefore, what we did
is we rewrote the equations
with the same bases.
Once we have the same
bases, we can just
set the powers
equal to each other.
Well, the problem with the
problems I chose is that you
cannot get the bases
to be the same.
If you look at
this first example,
I have 3 to the x equals 10.
Well, if you look
at the powers of 3,
you have 3 to the
first power is 3.
3 squared is 9.
3 cubed is 27.
So I can't rewrite
10 as a base 3.
And there is no other
way that I could do it so
that they have the same base.
So therefore, I have
to use my calculator.
Now, there's a couple
different mathematical ways
that we're going to do this.
And I'll show you--
I'll kind of work
through each and every one.
The first way that I
just prefer to do it
is just by rewriting it
into a logarithmic equation.
If you remember, if I
have b to the y equals x,
I can rewrite that as log
base b of x equal to y.
So if I have an equation
that's in exponential form,
I can always rewrite
it in logarithmic form.
So by taking this
equation and rewriting it
in logarithmic form, I
have log base 3-- remember,
the bases are always the
same-- of 10 equals x.
Now to plug this
in my calculator,
since I don't have
this as a base 10--
did I choose any of
them to have base 10?
No I didn't.
So what I'll have to do then is
use the change of base formula.
So I'll do log of 10
over log of 3 equals x.
So basically what I'll simply
do is just take my calculator.
And you can use
log or natural log.
It doesn't really matter.
I'll just do log of 10
divided by log of 3.
And that gives me
approximately 2.09.
I'm just going around
this to the nearest tenth.
So that's going to be
2.1 is approximately x.
I'll just write
it the other way.
x equals approximately 2.1.
OK, now when rewriting
them in exponential form,
it is very important to make
sure that we isolate this.
It has to be exactly
in this format.
It can't be in any of
these crazy formats.
So a lot of times
what we need to do
is what we call
isolate the exponent.
So you can see our exponent
here is 4 to the x.
So to isolate it, I see
that 2 is be multiplied.
So in this problem, what
I'm simply going to do
is divide by 2 on both sides.
So I have 4 to the x equals 10.
Again, I can't simplify.
I can't rewrite 4 and
10 with the same base.
Because 4 to the
first power is 4.
4 squared is 16.
4 cubed is going to be 64.
So therefore, again,
I can just rewrite
this in exponential form.
And the other way
to also do it is
it doesn't matter
if you're doing log.
You can also do the natural log.
And just try to do this.
Try doing both of them
and see that the answer is
going to be exactly the same.
Just make sure when you're
typing this in your calculator
you do an ln of 10 close
parenthesis divided by ln
of 4 and close parenthesis.
And this gives me approximately
1.7 as I round it.
OK, now, that's the
way that I prefer.
There's also another
mathematical rule
that we could use, which is--
I should probably do this.
Here's the change of base
formula-- log of base b
to a is equal to log of
a over log of b, which is
equal to ln of a over ln of e.
So it doesn't matter if
you use log or natural log.
The other form that
I wanted to show you
is if you have log base
b of a to the x-- oops--
base b of b raised to the
x, that's just equal to x.
And that's the way that a lot
of text books like to use.
And I'll do that as well
for this next example.
So the next example,
again, you can
see my exponent is 6 to the x.
I need to isolate my exponent.
So therefore, I could rewrite
it in exponential form.
So I'll subtract 10.
And then I have 6
to the x equals 37.
Now a lot of textbooks,
what they like to do
is have you get
rid of the exponent
by using the rules
of logarithms.
If you have a log of base
b of b raised to the x,
then it just equals x.
So therefore, what I can do is I
can take the log of both sides.
So it's log.
And I'm going to want to take
the log of base 6 0 6 to the x
equals log base 6 of 37.
Well, log base 6
of to the x is just
equal to x, which is log
base 6 of 37, which now I
can solve using my
one to one property--
or change of base formula.
I'm not going to
write it in here.
I'm just going to evaluate it.
Divided by log of 6,
which is approximately 2.
OK, so the next one
just has, again,
some more interesting
identities.
Again, the main important thing
is isolating the exponent.
I always just like to convert.
I don't really like using
the rules of logarithms,
or at least that
rule of logarithms.
That one is confusing
for a lot of students.
And it's just as simple
just to reconvert it
to logarithmic form.
However, the main
important thing,
though, if you want to use
the rules of logarithms
or just convert it to
exponential or logarithmic,
you have to have your
exponent isolated.
So in this case, I need to
treat this like a variable.
I need to treat the
exponent like a variable.
So how would I solve
1/2x plus 2 equals 8?
Well, to do that, I'm going to
subtract the 2 on both sides.
I have 1/2 4 to the
x minus 5 equals 6.
And I divide by 1/2, which is
the same thing as multiplying
by the reciprocal.
So now I have 4 to the x
minus 5 is equal to 12.
Now I can rewrite this
in exponential form.
So therefore, it's going
to be log base 4 of 12
equals x minus 5.
Then I'll add 5, add 5.
So x equals log
base 4 of 12 plus 5.
So therefore, now, to
approximate my answer,
I'm just going to do the log
of 12 end parenthesis divided
by the log of 4 end parenthesis.
And I'm going to take that
answer and add 5 to it.
And I have x equals
approximately 6.8
as I round it.
OK, so that's with numbers.
And again, those are the
easiest ones to understand.
It's either you're using
the one to one property
or using your calculator.
And the main important thing
is you could, actually,
for all the ones, even with
the one to one property,
you can still use your
calculator for all those.
But typically, using
the one to one property
is going to be much
faster and easier
than having to go back
and use your calculator.
Now what we're going to be
doing is going back into base e.
Now remember, base
e, we're always
going to be using our
natural logarithm.
So even though
the change of base
formulas you could have
used a log or natural log,
when we have a
base e, we're going
to have to use the
natural log to evaluate.
So in this case, again,
what you could do
is take the natural
log of both sides.
Because remember, the
natural log is base e.
So that's e to the x minus
1 equals ln base e of 12.
And that's why it's kind
of helpful of converting it
to exponential form.
That's why it's nice to think
about it using this property.
Because when you have base e,
the easiest thing to do is just
take ln of both sides.
Because now I just
have x minus 1
is equal to-- I don't need
to write base e-- ln of 12.
So therefore, I'll
just add 1, add 1,
x equals ln of 12 plus 1.
So now, instead of using the
change of base formula, which
I had to for my logarithms,
I can simply just do ln--
because it is base e-- of 12
and then add 1 to that result.
And I get 3.5, or x
is approximately 3.5
as I round to the nearest tenth.
But just like in
our other problems,
we have to make sure we
isolate our exponent.
So here I have e to the 7x.
But it doesn't matter.
That's still the
power of my exponent.
So the only thing I have
to do here is divide by 9.
So I have e to
the 7x equals 2/9.
Convert this to take
the ln of both sides.
Don't need to write ln e.
I just want to do that
so you guys remember
that it has a base e.
So therefore, I
have 7x equals ln
of 2/9 divided by
7 divided by 7.
So x equals ln of
2/9 divided by 7.
So now in my
calculator, I'm simply
just going to do ln of 2
divided by 9 end parenthesis.
And then we'll take that
answer and divide it by 7.
And I get negative 0.2, or
approximately negative 0.2.
OK, in the next one, again, we
have a little bit more crazier
power.
But again, it doesn't
really matter.
The main important
thing is you just
want that exponent to
be isolated, right?
Just isolate that exponent.
So you subtract 1 on both
sides because that's not
in the power.
So I have e to the 3x
minus 5 equals negative 3.
Now again, I did the
ln of both sides.
But again, ladies
and gentlemen, you
can convert this to exponential
form if you wanted to.
And even if you wanted
to convert it to log,
I mean, let's just do it.
It'd be log base e of
negative 3 equals 3x minus 5.
Well again, log base e is ln.
So it's ln of
negative 3 equals 3x
minus 5, which if
you just would have
taken the ln of both
sides, you would
have got the exact same result.
It just would have been
that ln on both sides
would've given you
the exact same thing.
So it's just different ways of
getting to the root product--
to getting to the answer.
You could take the log or
natural log of both sides.
Or you can just convert
to exponential form,
however way you feel
comfortable with.
Now we need to solve for x.
So I'm going to add 5.
So I have ln of negative
3 plus 5 equals 3x.
Then I'll divide by 3.
So my final is x equals ln of
negative 3 plus 5 divide by 3.
And that, approximately, is
ln of negative 3 plus-- oops.
Oh, can't do that.
Ah, your results cannot
equal a negative number.
That is correct.
e to the x equals negative 3.
Yes, you cannot have
ln equals negative 3.
I totally forgot about
that on this problem.
So remember, if you
look at our graph
of an exponential
equation, graph
of an exponential equation looks
something like this, right?
Well, if we're looking
for a variable,
you notice that
none of the y values
are going to be negative, right?
So we can't take e to
the x and have it equal.
So when we isolate it--
sorry, forgot about this--
when we isolate e
to the negative x,
we can't have it equal
to a negative power.
So therefore, in this
case, there is no solution.
So whenever an
exponential equation
is equal to a negative
number, therefore, we
know that there is no value x,
all these x values, whenever
we plug them into
equation, are always going
to give us positive answers.
So, for instance, if I had like
2 to the x equals negative 4,
there's no number you can raise
2 to to give you a negative 4.
There's no solution, right?
There is no number you can
take 2, raise it to a power,
and give you negative 4.
The most common response
I get from students
is 2 to the negative
second equals negative 4.
Well, that's not correct.
Because the 2 to
the negative second
is 1/4, which does
not equal negative 4.
OK?
So I kind of went along.
I was thinking to show you, if
you do the exact same thing,
you can convert it.
However, this is an example
of when you get into there,
you immediately can
just say no solution.
I kind of forgot I put
this problem in there.
So which brings us to our
next lovely problem, which
is in an odd form, nothing
like we have seen before.
And again, we are
trying to-- oh wait.
I didn't write what
that was equal to.
That's equal to 0.
My apologies.
It's an equation,
not an expression.
OK, so the first thing
I kind of noticed
is this kind of looks
like a trinomial.
Well, it is a trinomial.
It has three terms.
But it kind of looks
like a quadratic.
So what I'm going to do
is I'm going to replace--
I'm going to try to
rewrite this in terms
of my quadratic equations.
And what I'll do
is I'll write this
as x squared minus 4x minus 5.
So what I'm basically saying
is x is equal to e to the x.
OK?
This is what we call
our substitution method.
Now, if I was going to
do this in this case,
I would basically say do you
see how if I replace x with e
to the x I get that?
Now remember, the power
rule works in here
to give you e to that 2x.
Because remember,
you're adding them.
OK, so if I wanted
to solve this,
I would have to factor in
using zero product property.
Well, by factoring this
out, I get x minus 4 times
x plus 1-- no, x
minus 5-- equals 0.
Then, what I could do is use
the zero product property.
And I'd say x minus 5 equals 0.
And x plus 1 equals 0.
OK, well, I factored
it using quadratic.
And that's what I like to do.
But now I need to kind of
rewrite and plug everything
back in.
If I'm saying x is equal to
e to the x, then basically
what I'm saying is e to the x
minus 5 times e to the x plus 1
equals 0.
Now, let's go back and double
check to make sure this works.
What's e to the x
times e to the x?
e to the x times e to
the x is equal to e
to the x plus x, right?
Because when you
multiply exponents,
you add the powers, which
equal to e to the 2x.
OK?
So that's how I got that.
Negative 5 times
1 is negative 5.
And then negative 5 times
e to the x is negative 5x.
And 1 times e to the
x is 1e to the x.
Well, negative 5 e to
the x plus 1 e to the x
is going to equal
negative 4 e to the x.
Perfect.
So good.
So now by using my
zero product property,
I can say e to the x minus
5 equals 0 and e to the x
plus 1 equals 0.
Well, when I solve, I get e
to the x equals negative 1.
And we know that an exponential,
which I explained over here,
cannot equal a negative number.
So that's not going to
produce an equation.
So I just have 5.
So I have e to the x equals 5.
Then, taking the ln of both
sides, I can say the ln of 5
is equal to x.
So now all I need do is go back
to my calculator and just type
in ln of 5.
And that equals 1.6.
So x is approximately 1.6.
So there you go,
ladies and gentlemen.
That is how you solve
exponential equation
by using your calculator.
Thanks.
