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.