Hi everyone!
Today we’re going to talk about how to use laws of logarithms to simplify a logarithmic
function.
To complete this problem, we’ll first identify the laws we need to use and then apply them
to our logarithmic function in order to simplify it.
Let’s take a look.
In this particular problem, we’ve been asked to use laws of logarithms to evaluate log
base two of eighty minus log base two of five and we’re going to be using two laws of
logarithms here, one is log base a of x minus log base a of y can be converted into log
base a of x divided by y then the other one is log base a of x equal to y can be converted
into a to the y equals x.
So let’s just take this one step at a time and go through this problem, we’re trying
to simplify this original problem as much as we can.
So, we’re going to first equate our original problem to the first definition we have here.
We can see that when we have two logs with the same base, log base a and log base a of
x and y, and we have the same bases here, we have two and two with different x and y
where our x is equal to eighty and our y is equal to five, we can convert that into log
with the same base so we’ll still have base two but x in our numerator and y in our denominator.
The important thing to know about this definition is this negative sign here.
So if we had, for example, negative log base a of x plus log base a of y, the important
piece to realize is that whichever one of these is the negative log goes in the denominator,
over here, and whichever one is the positive goes in the numerator.
So, in our case, log base two of eighty is positive, this log base two of five is negative,
so, we’re going to put eighty in the numerator and five in the denominator.
So, our log base two of eighty, so log base two of eighty, minus log base two of five
will become log base two of eighty over five.
Obviously, we can simplify that by dividing eighty by five and we’ll get log base two
of sixteen because eighty divided by five is sixteen.
So now, we’ve got something that’s much simpler than what we started with but we can
still simplify further.
So if we use our second rule here, that log base a of x is equal to y, something in that
form, can be converted to a to the y equals x.
All we have to do here is take our log base two of sixteen and let’s go ahead and set
that equal to… let’s call it y to keep it in the same format or as close to the same
format as our definition here.
So if we set that equal to some variable, in this case we’ll just call it y, then
we can transform our log base two of sixteen equals y into this second form here, a to
the y equals x.
So notice that the log base a will go away completely.
So notice that the log goes away, it’s nowhere to be found on the right hand side over here,
a to the y equals x.
But, we do have this base a, and that shows up over here on the right hand side, so our
base is two, so we’ll go ahead and write that two.
Now, we’re looking for the y over here and the y in this case is what’s over here on
the right hand side of our first equation.
In that case, for us, we’ve got y so we’ll just call this two to the y.
Now, over here on the right side, all we need is x which we know is what we’ve got over
here with our log, and, in our case, that’s sixteen.
So, we’ll call this two to the y equals sixteen.
And that’s the way that we can simplify log base two of sixteen equal to some variable,
it turns into two to the variable is equal to sixteen.
And now, of course, at this point, we know that the y is just equal to four, because
we know that two to the four is equal to sixteen, which means that our final answer here is
just four.
We can simplify our original problem, log base two of eighty minus log base two of five,
all the way down to just four.
And that’s it.
That’s one example of how to use the laws of logarithms to simplify a problem like this
one.
So, I hope you found that video helpful.
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