In the last section we looked at some expressions
like, “What is the third root of twenty-seven, squared?” The math is kind of ugly looking.
The procedures are clunky and it is very easy to lose
sight of the objective. What this
expression is asking is what number cubed is twenty-seven squared. You could always square the 27, to arrive at
729 and see if that is a perfect cube.
There is a much more elegant way to go about this type
of calculation. Turns out if we rewrite
this expression with a rational exponent, life gets easier.
These two statements are the same. They ask the same question, what number cubed
is twenty-seven squared?
By now you should be familiar with perfect cubes and
squares. Hopefully you’re also familiar
with higher powers of 2 and 3, as well as a few others. For example, you should recognize that 625 is
54. If you don’t know that yet, a cheat sheet
might be helpful.
Let’s look at our expression again. If you notice that 27 is a perfect cube, then
you can rewrite it like this:
Maybe you see what’s going to happen next, but if not,
we have a power raised to another here, we can multiply those exponents. Three times two-thirds is two. This becomes three squared.
Not too bad! We
factor, writing the base of twenty-seven as an exponent with a power that
matches the denominator of the other exponent, multiply, reduce, done!
Let’s look at another.
We mentioned earlier that 625 was a power of 5, the
fourth power of five. That’s the key to
making these simple. Let’s rewrite 625
as a power of five.
We can multiply those exponents, giving us five-cubed,
or 125. Much cleaner than finding the
fourth root of six hundred and twenty-five cubed.
What about something that doesn’t work out so, well,
pretty? Something where the base cannot
be rewritten as an exponent that matches the denominator?
This is where proficiency and familiarity with powers
of two comes to play. Thirty-two is a
power of two, just not the fourth power, but the fifth.
If we multiplied these exponents together we end up
with something that isn’t so pretty, 215/4. We could rewrite this by simplifying the
exponent, but there’s a better way.
Consider the following, and note that we broke the five twos into a
group of four and another group of one.
Now we’d have to multiply the exponents inside the
parenthesis by ¾,
and will arrive at:
Notice that 23/4 is irrational, so not much we can do with it,
but two cubed is eight. Let’s write the
rational number first, and rewrite that irrational number as a radical
823−−√4, or 88–√4.
There’s an even easier way to think about these rational exponents. I’d like to introduce something called
Logarithmic Counting. For those who don’t know what logarithms are, that
might sound scary.
Do you remember learning how to multiply by 5s…how you’d skip count?
(5, 10, 15, 20, …) Logarithmic counting is the same way, except
with exponents. For example, by 2: 2, 4, 8, 16, 32, … Well, what’s the fourth step of 2 when
logarithmically counting? It’s 16,
Let’s look at 163/4. See the denominator of four? That means we’re looking for a fourth root, a
number times itself four times that equals 16.
The three, in the numerator, it says, what number is three of the four
steps on the way to sixteen?
Above is how we get to sixteen by multiplying a number
by itself four times. Do you see the
third step is eight?
Let’s see how our procedure looks:
(24)3/4=241×34=23, or 8.
24−−√4×24−−√4×24−−√4=2×2×2, or 8.
The most elegant way is to realize the 16 is the
fourth power of 2, and the fraction ¾ is asking us for the third entry. What is 3/4s
of the way to 16 when multiplying (exponents)?
Let’s look at 6252/3. Let’s do this three ways, first with radical
notation, then by evaluating the base and simplifying the exponents, and then
by thinking about what is two thirds of the way to 625.
this is going to be a tricky problem because 625 is NOT a perfect cube. It is the fourth power of 5, though, which
means that 125 (which is five-cubed) times five is 625.
53×53×52−−−−−−−−−−√3=5×525−−√3, or 2525−−√3
52×52/3=25×52/3, or 2525−−√3
A little better, but still a few sticky points.
Now our third method.
6252/3 asks, “What is two thirds of the way to 625,
for a cubed number?”
This 625 isn’t cubed, but a factor of it is.
This could also be written as:
I am certain that 5 to the two-thirds power is
irrational because, well, five is a prime number. Let’s deal with the other portion.
steps to 125 are: 5 25
second step is 25.
To summarize the denominator of the rational exponent
is the index of a radical expression.
The numerator is an exponent for the base. How you tackle the expressions is entirely up
to you, but I would suggest proficiency in multiple methods as sometimes the
math lends itself nicely to one method but not another.
Simplify the following:1. (16x16)3/42. 1285/63. 1253−−−−√54. 323/55. (81x27)2/3
Select Practice Problems Review
1. (16x16)3/4 A key piece of information that can make this
easy is that the base, 16, is a perfect 4th power. Four is the denominator of the exponent. So what we will do is rewrite 24
in place of 16.
Inside the parenthesis are two bases, the 2 and the x.
Both bases get their exponents multiplied by ¾.
3. 1253−−−−√5 This is a tricky problem because 125 is not a
5th power! The first thing
I’d suggest is rewriting the radical notation as an exponent, and then writing
125 as 53.
Here we need to be
delicate. When we multiply 3 and 3/5, we
get 9/5. This is 1 and 4/5.
You could leave the
answer as 5 × 54/5, or rewrite it in radical notation.
1253−−−−√5=554−−√5 or 5×54/5