Professor Dave here, let’s solve some equations.
Now that we know what logarithms are, we are
ready to solve exponential and logarithmic
equations.
Take something like two to the X equals sixteen.
It’s true that we could just intuitively
understand that X must be four, because we
can see that two must be raised to the fourth
power for this equation to be valid.
But things get trickier when this number isn’t
so obvious.
What about two to the X equals seventeen?
If you’re not a computer, you can’t do
this in your head, so we have to use logarithms.
Taking the log base two of both sides gives
us this.
Now it follows that two to this power equals
two to the X power, so X must be equal to
log base two of seventeen.
This may not be aesthetically pleasing, but
it is the precise solution for this equation.
If we wanted to get an answer that can be
plugged into a calculator, we could instead
take two to the X equals seventeen and just
take the natural log of both sides.
Then, we could drag this X to the front and
get X times the natural log of two equals
the natural log of seventeen.
Dividing by the natural log of two, there’s
our answer in terms of natural logs, which
is actually what we would have gotten if we
had applied the change-of-base property to
our original answer to get it into natural
log form.
Now let’s look at what happens when the
exponent becomes more complicated.
What about three to the quantity two X minus
seven equals twenty-seven.
Again, rather than guess and check, let’s
solve the equation.
We notice that we have this base three being
raised to the exponent.
Can we express the other side of the equation
in a similar way?
In fact we can, we can make it three cubed.
Now all we have to do is recognize that if
three raised to this power equals three raised
to the third power, then these exponential
terms must be equal to each other.
Another way of rationalizing this is by taking
the log base three of both sides.
The bases all cancel, leaving us with two
X minus seven equals three.
From here it’s trivial, we add seven and
divide by two to get X equals five.
Sometimes changing the base will manipulate
an existing exponent.
Say we have two to the X plus three equals
four to the X minus one.
Here, to get the same base, we have to change
four to two squared, which means we have two
raised to the second power raised to the X
minus one power.
But when this happens, we know that the exponents
simply multiply, so we have two to the two
times quantity X minus one, or two to the
two X minus two.
Now that we have the same base on both sides,
we could take the log base two of both sides,
or don’t even bother and just recognize
that the exponents must be equal to each other.
Add two, subtract X, and X must be five.
Now let’s see what happens when we can’t
get the same base on both sides.
What about three to the X plus two equals
two to the X minus one.
We can’t get this in a form where logs will
cancel both bases, we could only get one and
not the other.
So instead, let’s take the natural log of
both sides.
Once we’ve done that, we can bring these
exponents to the front.
Then, we can distribute each log over the
adjacent binomial.
Let’s then put the terms with X on one side
and the terms without X on the other side,
that way we can factor out an X.
Then we just divide both sides by this term,
and we have X in terms of natural logs that
we could plug into a calculator, which will
allow us to get an approximate value for X.
We could also continue to condense this expression
if that was desirable for one reason or another.
Let’s quickly go over some logarithmic equations
we may want to solve.
Take log base two of X minus four equals three.
This is easy to solve if we know the definition
of a logarithm.
Two cubed must be equal to X minus four, so
X minus four equals eight, and X equals twelve.
That’s quite straightforward, but sometimes
we might have to use properties of logs in
our solutions.
How about log base two of X plus log base
two of X minus three equals two.
Well these logs have the same base, so we
can use the product rule to condense them,
and get log base two of X times X minus three,
or X squared minus three X.
Now by the definition of a logarithm, two
squared, or four, must equal X squared minus
three X.
Now we just have a polynomial to solve.
Let’s get four on the other side, and if
we stare at this for a few seconds we will
see that we can factor this into X minus four
and X plus one, making four and negative one
the solutions to this equation.
You might find ones that are trickier than
these, but as long as you use all the properties
of logs that you know, you should do just
fine, so let’s check comprehension.
