### Video Transcript

In this video, our topic is
rearranging formulas involving logarithms. By the end of the lesson, weβll
understand just how it is that the left-hand side of this equation we see on screen,
as long and as complicated as it looks, really does equal simply one.

As we get started working with
logarithms, letβs consider this expression here, π is equal to two raised to the
π. Given this expression, we know that
if we take the log base two of this number π, then the result weβll get is π. But then notice this. Since π is equal to two raised to
the π, we could substitute that in for π in our logarithm equation. And that would give us this
expression, the log base two of two raised to the π equals π. Here, we have an exponential
function two raised to the π, and this is the argument of a log base two
function. And we see that with the base of
our logarithm and the base of our exponent being the same, the result of this whole
calculation on the left is simply the exponent all by itself.

We could say then that our
logarithm and our exponent are canceling one another out, that theyβre inverses. And it turns out that this is true
in general, regardless of the base that weβre working with. Mathematically, we can express that
fact this way. We can say that a logarithm
function with a given base π₯, when this is applied to an exponential function with
a base π₯ and an exponent π¦, will invert that function. And the result will be the exponent
π¦. Notice that this example here, with
a base of two for the log and the exponent, is a special case of this general
relation. Itβs important to note that, for
this relation to hold, the base of our exponential function and the base of our
logarithmic function must be the same.

We can use this fact to solve for
certain variables in mathematical equations. For example, say that we were given
this equation here and we wanted to make the variable π¦ the subject. Given this task, we might first
think to divide both sides of the equation by π, whatever that value is. That would give us this result. But now, how do we bring π¦
downstairs, so to speak, out of the exponent of this function? We can recall that because
logarithms and exponents are inverses, then if we take the logarithm, base π, of
this exponent, weβll wind up with just that exponent π¦. Notice that in order to apply this
logarithm here and maintain our mathematical equality, we have to apply it to the
other side of our equation too.

So then, by this rule here, the log
base π of π to the y simplifies to π¦. And now we have this variable as
the subject of our equation. We start to see then that
logarithms as well as exponents, since theyβre inverses, can be useful tools for
helping us rearrange formulas. To see another example of how this
might work, letβs consider this equation here, π is equal to π raised to the π
plus π‘. We can recall that this particular
exponential function, this letter π raised to a value, is called the exponential
function, where π has a numerical value of approximately 2.71. So then, given this equation, how
would we make π , the exponent of this exponential function, the subject?

The first thing we might think to
do is to subtract π‘ from both sides so that it cancels on the right side. But then, what can we do next to
bring π downstairs and make it the subject? Recalling once again that
logarithms and exponents are inverses, we can realize that the exponential function,
which is an exponent with a particular value for its base, has as its inverse a
particular logarithmic function called the natural logarithm.

The fact that these two functions,
the exponential function and the natural logarithm, are inverses means that
according to our rule here, if we were to replace π₯ to the π¦ with π to the π₯ and
the logarithm base π₯ with the natural logarithm, then the left-hand side of our
expression would look like this, the natural logarithm of π to the π₯. And this is equal to the exponent
π₯.

When we say then that the
exponential function and the natural logarithm are inverses, this is part of what
that means mathematically. We say part of what it means
because really we could write this in the opposite order. Here we have the natural logarithm
of an exponential function. But we could equivalently take an
exponential function and have as its exponent the natural logarithm of π₯. And once again because these two
functions are inverses of one another, this is equal to π₯.

All this to say when we come back
over here trying to make π our subject, if we apply the natural logarithm to both
sides of the expression, that looks like this. And then, since the natural
logarithm of π raised to the π₯ is equal simply to π₯, we can then say that the
natural logarithm of π to the π simplifies to π itself. As we work with these expressions
involving exponents and logarithms, itβs worthwhile to recall some rules for working
with those functions.

Starting with exponents, we can
remember that if we have some number raised to the power π₯ multiplied by that same
number raised to a different power, π¦, then that product is equal to that base,
that particular number weβve called π, to the π₯ plus π¦ power. And then, if we consider that the
opposite of multiplication is division and the opposite of addition is subtraction,
we come to understand the second rule for working with exponents. And then lastly, if we have some
base π raised to the power π₯ and that whole expression is raised to the power π¦,
then thatβs equal to that base π to the π₯ times π¦ power.

Along with these rules for
exponents, we can recall some corresponding ones for logarithms. First off, that a log with a given
base of a product of two numbers, π₯ times π¦, is equal to the sum of the log with
that same base of those two separate values, π₯ and π¦. Similarly, the log base π of a
fraction, π₯ divided by π¦, is equal to the log base π of the numerator minus the
log base π of the denominator. And lastly, if we have a logarithm
of an exponent and notice that in this general case the base of our log is not the
same as the base of our exponent, then this is equal to that exponent times the log
base π of the base of the exponential function π₯.

We can see in this last rule here a
general expression that we found a specific expression for over here. That is, when the base of our
logarithm and the base of our exponential function are equal, then this expression
on the left simplifies to the exponent itself. And over here, in our very general
rule, if we were taking the log base π of π₯, where π₯ was equal to π, then all of
this would be equal to one. And we would find a result that
agrees with this expression here. Now, our main focus here is not the
rules themselves, but rather to let them help us rearrange formulas involving
logarithms and exponents.

Knowing all this, letβs clear a bit
of space at the top of our screen and look once more at that expression we found on
the opening screen of this video. That expression looked like
this. And now we can analyze the
left-hand side to see if it really does equal what weβve said it equals. Weβll start at the most interior
mathematical function. Thatβs this exponential function
here, π raised to the natural log of one. Weβve seen that π raised to the
natural log of some number π₯ is equal to π₯ because the exponential function and
the natural logarithm are inverses. So that means that all of this is
equal to one.

Next, we can look at this
exponential function, which is π raised to the log base π of one. To see what this equals, letβs look
down here for a moment. We said that taking the natural
logarithm of an exponential function inverts those functions in the same way that
taking the exponent of a natural logarithm does. And we recall further that the
exponential function π and the natural logarithm ππ are simply an exponent and a
log with a particular base value. That tells us that this expression
here is a specific application of this more general rule.

In this rule, we see that weβre
first applying a logarithmic function to an exponential one. But since logarithms and exponents
are inverses, we could equally well do it the opposite way, use an exponential
function first and take as its argument a logarithm. Using these variables, that would
look like this: π₯, the base of our exponential function, raised to the log base π₯
of π¦. And this too is equal to π¦.

The important equality here is that
the base of our logarithmic function, π₯, is the same as the base of our exponential
one. When thatβs true, then these
equalities all hold. As we look at this part of our
expression, we can notice that it applies to π raised to the log base π of
one. Here, weβve got an exponential
expression with a base π equal to the base of our logarithm. And so what this all simplifies
down to is the value in the parentheses of that logarithm, that is, one.

And now weβll consider the natural
logarithm of π raised to the first power. Looking down at our rule here, the
natural logarithm of π raised to any number is equal to that number. So therefore, this is all one. And now on the left-hand side, we
have the logarithm base π of π to the one. If we apply this rule here, that
the log base π₯ of π₯ to the π¦ is equal to π¦, then we see that all of this
expression will simplify down to the exponent here. And so all of that left-hand side
really does equal one. Letβs get a bit more practice now
with these ideas through an example exercise.

The quantities π, π, and π are
related to each other by the formula π is equal to π times π raised to the
π. Which of the following shows a
rearrangement of this formula with π as the subject? (A) π is equal to the natural
logarithm of π minus π. (B) π is equal to π raised to the
π divided by π. (C) π is equal to the natural
logarithm of π divided by π. (D) π is equal to the natural
logarithm of π divided by π. (E) π is equal to π raised to the
π divided by π power.

So given these quantities π, π,
and π, which are related to one another through this formula, what we want to do is
rearrange the formula so that π is the subject and see which one of our answer
options that rearrangement agrees with. Our mission, then, is to rearrange
this expression so it reads π is equal to some other quantities. The first step we can take to do
that is to divide both sides of the equation by the quantity π. This cancels that term out on the
right. Our expression then is in this
form. But then, how do we bring π down
from this exponent?

We can do this by recalling that
the exponential function, π raised to some value, has a corresponding inverse
function, that is, a function that undoes the exponential operation. This inverse function is called the
natural logarithm. The way it works, if we have an
exponential function of some variable π₯, if we take the natural logarithm of that
exponential function, then the result is the exponent π₯. In our example, the exponent weβre
interested in is π. So if weβd apply the natural
logarithm to π raised to the π, then this rule tells us we can isolate π.

An important thing to remember
though about applying the natural logarithm function is that itβs an operation
thatβs like multiplication or division. In other words, if we apply the
natural logarithm to one side of our equation, then we need to apply it to the other
side as well. So thatβs what weβll do. Here, weβve applied that natural
logarithm to both sides, and on the right, we have the natural log of π to the
π. As we said, by applying this rule
here, this right-hand side simplifies to the exponent itself. And that leaves us with this
expression here, π is equal to the natural logarithm of π divided by π. Scanning through our different
answer options, we see that option (D) agrees with this result.

So if we have three quantities π,
π, and π related to one another according to this equation here, then a
rearrangement of this formula with π as the subject is π is equal to the natural
logarithm of π divided by π.

Letβs summarize now what weβve
learned about rearranging formulas involving logarithms. In this lesson, we learned that
logarithms and exponents are inverse functions. Mathematically, this means that the
logarithm base π of π raised to the π₯ power is equal to π₯ and, equivalently, so
is π raised to the log base π of π₯. We saw further that a specific case
of this involves the exponential function, π to the π₯, and the natural logarithm,
ππ of π₯.

And lastly, we saw that when
theyβre applied to mathematical equations, logarithms and exponents operate on all
terms on both sides of an equation. In this way, applying a logarithmic
or exponential function is like multiplying by a constant value all through an
equation. Just as that multiple must be
applied to every term, so must the logarithm or exponent applied. This is a summary of rearranging
formulas involving logarithms.