Video Transcript
In this lesson, we’ll learn how to
solve exponential equations using the properties of exponents. We are particularly interested in
the rules for working with exponents when finding the product, quotient, or power in
addition to how we evaluate zero, negative, and fractional exponents. So let’s recall these first. The first rule is the rule for
multiplying numbers with exponents. Note that this only works if the
base, that’s the big number which here is the 𝑥, is the same. If this is the case, when we
multiply these terms, we add their exponents. So, 𝑥 to the power of 𝑎 times 𝑥
to the power of 𝑏 is 𝑥 to the power of 𝑎 plus 𝑏.
Similarly, if we divide these types
of numbers, we subtract their exponents. 𝑥 to the power of 𝑎 divided by 𝑥
to the power of 𝑏 is 𝑥 to the power of 𝑎 minus 𝑏. When we’re working with
parentheses, in other words, we’re raising an exponential term to another exponent,
we multiply these. So, 𝑥 to the power of 𝑎 to the
power of 𝑏 is the same as 𝑥 to the power of 𝑎 times 𝑏. We might also recall that anything
to the power of zero is one. A negative exponent tells us to
find the reciprocal, so 𝑥 to the power of negative 𝑎 is one over 𝑥 to the power
of 𝑎. And a fractional power tells us to
find a root, so 𝑥 to the power of one over 𝑎 is the same as the 𝑎th root of
𝑥.
Now, in this lesson, we’re going to
use these rules to solve equations involving exponents. And the best way to see how this
might look is to consider an example.
Given that two to the 𝑥th power is
equal to 32, find the value of 𝑥.
Here we have an equation involving
an exponent which is a variable. The exponent here is equal to
𝑥. Now, usually, to solve an equation,
we’d look to perform a series of inverse operations, but the inverse to finding the
𝑥th power is to find the 𝑥th root, which doesn’t really help us much. Instead, it’s worth noticing that
32 can be written as a power of two. In fact, we know that two to the
fifth power is equal to 32. And this means we can rewrite our
equation as two to the 𝑥th power equals two to the fifth power.
So, how does this help? Well, now the base — that’s the big
number; here that’s two — is the same. And so we can say that for this
equation to make sense, the exponents must also be the same. That is, 𝑥 is equal to five. Now, whenever we’re solving
equations, it always makes sense to check our answer by substituting it back into
the original equation. Let’s let 𝑥 be equal to five. And then two to the 𝑥th power
becomes two to the fifth power, which is indeed equal to 32. And so, given that two to the 𝑥th
power is equal to 32, 𝑥 must be equal to five.
So, that’s a fairly straightforward
example of how to solve exponential equations. What we’re going to do next is
incorporate some of the rules of exponents in an example.
Given that eight to the power of 𝑦
equals four to the power of 𝑧 which equals 64, find the value of 𝑦 plus 𝑧.
So this is a rather unusual-looking
equation because it has three parts to it. Now, the key to answering this
problem is to spot that each of the numerical parts of our equation can be written
as a power of some common number. In fact, they can be written as two
to the power of something. The first few powers of two are
shown. We know two squared is four, two
cubed is eight, all the way up to two to the sixth power is equal to 64. So, let’s replace each numerical
part of our equation with its associated power of two. Eight is two cubed. So, eight to the power of 𝑦 is two
cubed to the power of 𝑦. Then, four is two squared, so four
to the power of 𝑧 is two squared to the power of 𝑧. And finally, 64 is two to the sixth
power.
But how does this help us? Well, if we recall a rule for
working with powers of exponents, we know that 𝑥 to the power of 𝑎 to the power of
𝑏 is equal to 𝑥 the power of 𝑎 times 𝑏. Here we multiply the exponents. And so, two cubed to the power of
𝑦 can be written as two to the power of three times 𝑦 or just two to the power of
three 𝑦. And then two squared to the power
of 𝑧 is the same as two to the power of two 𝑧. And so, our equation is now two to
the power of three 𝑦 equals two to the power of two 𝑧 which equals two to the
power of six.
Now that the base, that’s the big
number which is here two, is the same, we can say that for this equation to make
sense, the exponents themselves must also be the same. In other words, three 𝑦 must be
equal to two 𝑧, which must in turn be equal to six. Now, we can actually split this
equation up to find individually the value of 𝑦 and 𝑧. Let’s begin by equating three 𝑦 to
six. Three 𝑦 equals six. And so we’ll solve this equation
for 𝑦 by dividing both sides by three. Three 𝑦 divided by three is 𝑦,
and six divided by three is two. So we can say that 𝑦 must be equal
to two.
Similarly, we can equate the other
two parts of the equation. We can say that two 𝑧 is equal to
six. And then to solve for 𝑧, we simply
divide through by two. Two 𝑧 divided by two is 𝑧, and
six divided by two is three. So we found that 𝑦 is equal to two
and 𝑧 is equal to three. That’s not enough, though. We needed to find the value of 𝑦
plus 𝑧. Well, we can now say that that must
be equal to two plus three, which is of course equal to five. And so, given the restrictions on
our equation, we can say that 𝑦 plus 𝑧 must be equal to five.
We’re now going to consider how to
find the solution set of an exponential equation when the exponents are
binomials.
Find the value of 𝑥 for which
eight to the power of 𝑥 plus two equals two to the power of 𝑥 plus four. Give your answer to the nearest
tenth.
Here we have an equation that’s
made up of two exponentials, but where the exponentials are binomials; they’ve got
two terms in them. And in fact, the key to answering
this problem is in spotting that each of the bases of our equation — that’s the big
number, so eight and two — can be written as a power of some common number. In fact, if we recall that two
cubed is equal to eight, we can rewrite the base on the left-hand side as two
cubed. So, let’s see what that looks
like. We get eight to the power of 𝑥
plus two equals two cubed to the power of 𝑥 plus two. And that’s actually really useful
because we know that we can multiply the exponents now.
We recall that 𝑥 to the power of
𝑎 to the power of 𝑏 is 𝑥 to the power of 𝑎 times 𝑏. And so this becomes two to the
power of three times 𝑥 plus two. Let’s replace this expression in
our earlier equation. When we do, we find that two to the
power of three times 𝑥 plus two is equal to two to the power of 𝑥 plus four. So, how does this help? Well, now that the base is the same
on both sides — it’s two — we can say that for the equation to make sense, the
exponents themselves must also be the same. That is, three times 𝑥 plus two is
equal to 𝑥 plus four.
To solve this equation for 𝑥,
we’re going to begin by distributing the parentheses on the left-hand side. And we do so by making sure that we
multiply three by the 𝑥 and then three by the two. Three times 𝑥 is three 𝑥, and
three times two is six. So, our equation is three 𝑥 plus
six equals 𝑥 plus four. Since we’re looking to isolate the
𝑥, we’re going to begin by subtracting the smaller number of 𝑥. So we’re going to subtract 𝑥 from
both sides. Three 𝑥 minus 𝑥 is two 𝑥, and 𝑥
minus 𝑥 is zero. So, our equation becomes two 𝑥
plus six equals four. Next, we subtract six from both
sides. Two 𝑥 plus six minus six is two
𝑥, and four minus six is negative two.
Our final step to solve for 𝑥 is
to divide through by two. And when we do, we find that 𝑥 is
equal to negative one. Now of course, negative one is an
integer solution, so we don’t need to do any rounding despite the fact that it tells
us to in the question. What we can do, though, is check
our solution is definitely right by substituting it back into the original
equation. Eight to the power of 𝑥 plus two
becomes eight to the power of negative one plus two. That’s eight to the power of one,
which is eight. Then, two to the power of 𝑥 plus
four becomes two to the power of negative one plus four. That’s two cubed, which is also
equal to eight. And since these expressions are
equal, we can say our answer is correct. 𝑥 is equal to negative one.
In our next example, we’ll look at
how to solve an exponential equation that involves some absolute value.
Find the solution set of two to the
power of the absolute value of eight 𝑥 minus 12 equals eight to the power of four
𝑥 minus four.
The key to solving this equation is
to spot that we can rewrite eight as two cubed, thereby creating an equation where
the bases are equal. With the expression we have on the
right-hand side, if we replace eight with two cubed, we get two cubed to the power
of four 𝑥 minus four. But then, of course, one of our
rules for working with exponents can help us to simplify this. We know 𝑥 to the power of 𝑎 to
the power of 𝑏 is the same as 𝑥 of the power of 𝑎 times 𝑏. When we’re dealing with
parentheses, we can multiply the exponents, and so we can rewrite this as two to the
power of three times four 𝑥 minus four.
And so then, our original equation
can be rewritten as two to the power of the absolute value of eight 𝑥 minus 12
equals two to the power of three times four 𝑥 minus four. Now, this is really useful because
now that the bases are the same — it’s two on both sides — we can say that for this
equation to make sense, for everything to be equal, their exponents must themselves
be equal. That is, the absolute value of
eight 𝑥 minus 12 must be equal to three times four 𝑥 minus four.
Let’s distribute these parentheses
by multiplying each term inside the parentheses by the three on the outside. And when we do, we get 12𝑥 minus
12. So, how do we solve an
absolute-value equation? Well, we know that the
absolute-value symbol takes any negative input and essentially makes it
positive. And so, what we do is we change the
sign of what our absolute value is equal to. So we say that either eight 𝑥
minus 12 is equal to 12𝑥 minus 12 or eight 𝑥 minus 12 is equal to the negative of
12𝑥 minus 12.
Now, of course, if we distribute
the parentheses on the right-hand side, we get negative 12𝑥 plus 12. So let’s solve both equations. With our first equation, we’re
going to subtract eight 𝑥 from both sides, and we get negative 12 equals four 𝑥
minus 12. Then we add 12 to both sides. And when we do, we get zero is
equal to four 𝑥. Well, the only way for this to be
the case is if 𝑥 itself is equal to zero. So, that’s one potential
solution. Let’s solve the second
equation. With our second equation, we’re
actually going to begin by adding 12𝑥 to both sides, so we get 20𝑥 minus 12 equals
12. Then we add 12 to both sides, and
we get 20𝑥 equals 24. And then we divide through by 20,
and we find 𝑥 is equal to 24 over 20, which simplifies to six-fifths.
What we’re going to do to check
both of these solutions is to substitute them back into our original equation and
check they both actually work. If we substitute zero into our
original equation, the left-hand side becomes two to the power of the absolute value
of negative 12 and the right becomes eight to the power of negative four. Then the absolute value of negative
12 is simply 12, and eight to the power of negative four is one over eight to the
power of four. Now, if we actually evaluate these,
we find two to the twelfth power is 4096. And then the right-hand side
becomes one over 4096. Now, we can actually see these are
not equal to one another. And so, actually, the solution 𝑥
equals zero doesn’t actually work. So we’re going to disregard that
solution.
Let’s repeat this process with the
second value of 𝑥. On the left-hand side, we get two
to the absolute value of negative 2.4, and that should be equal to eight to the
power of 0.8. But of course the absolute value of
negative 2.4 is just 2.4. So let’s evaluate both sides. When we do, we find that they are
both equal to roughly 5.27, so this solution does work. Now, of course, another way to
check this would have been to rewrite two to the power of 2.4 as two cubed to the
power of 0.8. And that’s because three times 0.8
is 2.4. Then we know that two cubed is
eight. So we’re actually saying eight to
the power of 0.8 is equal to eight to the power of 0.8.
Now, of course, the question was
asking us to find the solution set to our equation. And when there’s only one value, we
can use these sort of squiggly brackets to help us. The solution set is the set
containing the element six-fifths.
In our final example, we’re going
to look at how a bit of observation can help us to solve more complicated
exponential equations.
Determine the solution set of 𝑥 to
the power of 𝑥 squared minus 64 equals six to the power of 𝑥 squared minus 64.
Now let’s begin by noticing that
the exponents on each side of our equation are in fact equal. And so, for both sides of our
equation to be equal, one value of 𝑥 would be when the bases, the big numbers, are
equal. In other words, if we have 𝑥 is
equal to six, both sides of our equation are identical, so that is one solution. But are there any other
options? Well, another way to ensure that
two sides of our equation are equal is to have a power of zero since anything to the
power of zero is one. And so what we could do is say that
the exponent 𝑥 squared minus 64 is equal to zero.
Let’s solve for 𝑥 by adding 64 to
both sides to give us 𝑥 squared equals 64. And then, finally, we take the
square root of both sides, remembering, of course, to take both the positive and
negative square root of 64. But since the square root of 64 is
eight, we can say that the solutions to this equation are 𝑥 equals negative or
positive eight. And so, so far, we have three
possible values of 𝑥. But in fact, there is one more. And that solution is 𝑥 equals
negative six. So, why does 𝑥 equals negative six
work?
Well, imagine 𝑥 is equal to
negative six. When it is, the exponent becomes
negative six squared minus 64, which is negative 28. And we know that if 𝑎 is an even
number, the negative 𝑥 to the power of 𝑎 is the same as 𝑥 to the power of 𝑎. And so, the left-hand side is
negative six to the power of negative 28. But because negative 28 is even,
this is the same as six to the power of negative 28, which is what we get on the
right-hand side. And so, 𝑥 equals negative six also
has to be a solution. Now, of course, we only chose
negative six because we wanted to match the bases. We wouldn’t go ahead and choose any
value of 𝑥 that would make the exponents equal. It only works for negative six.
And so we can say the solution set
to our equation is the set containing the elements six, negative six, eight, and
negative eight.
In this video, we learned that we
can use the rules of exponents to help us solve exponential equations. We saw that if we can create a
common base by using these rules of exponents, we can then equate the exponents
themselves and solve as normal. We saw that we should always check
whether all answers given are actually valid by substituting them back into the
original equation and that if we’re working with negative bases, then we should
watch out for any possible even exponents that would create extra solutions.
