Try This First
Try this short activity before reading the lesson.
> Fold a sheet of paper in half, then in half again, and keep going until you cannot. After $n$ folds the stack is $2^n$ layers thick: $1$ fold gives $2$ layers, $2$ folds give $4$, $3$ folds give $8$. Now imagine a "half fold", done so that two half folds in a row make one full fold. How many layers would one half fold have to give? Is it a whole number?
The Big Idea
In Grade 8 you used the exponent laws on numbers and single letters: add the exponents to multiply, subtract them to divide, multiply them for a power of a power, and read a negative exponent as "one over". This chapter keeps every one of those laws and asks them to carry more weight. First, the base can be a whole product or quotient, like $(3x^2y)^4$ or $\left(\frac{2a^3}{b}\right)^{-2}$, and the exponent outside the parentheses reaches every factor inside. Second, the exponent can be a fraction.
The fraction case is forced, not invented. If the law $(a^m)^n = a^{mn}$ is to keep working, then $\left(9^{1/2}\right)^2 = 9^1 = 9$. So $9^{1/2}$ is a positive number whose square is $9$, which is $\sqrt9 = 3$. In the same way $a^{1/n} = \sqrt[n]{a}$ and $a^{m/n} = \left(\sqrt[n]{a}\right)^m$: the denominator names the root, the numerator names the power. The half fold from Try This First is $2^{1/2} = \sqrt2 \approx 1.41$ layers, which is why it could never be a whole number.
With both ideas in hand you can simplify $\left(8x^6\right)^{2/3}$, rewrite $\sqrt[3]{x^2}$ as $x^{2/3}$, solve $4^{x+1} = 8^x$ by writing both sides as powers of $2$, and read a growth formula like $N = 500 \cdot 2^{t/3}$ in which the exponent is itself a fraction of time.
Words to Know
- base, exponent: In $a^n$, $a$ is the base and $n$ is the exponent. In this course the base may be a product or quotient of variables, and the exponent may be any rational number.
- reciprocal: The number you multiply by to get $1$. A negative exponent asks for the reciprocal of the power: $\left(\frac{x}{3}\right)^{-2} = \left(\frac{3}{x}\right)^2 = \frac{9}{x^2}$.
- rational exponent: An exponent written as a fraction $\frac mn$. For $a > 0$, $a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$, so $x^{3/2} = \sqrt{x^3}$ and $x^{-1/3} = \frac{1}{\sqrt[3]{x}}$.
- scientific notation: A nonzero number written as $c \times 10^k$ with $1 \le |c| < 10$ and $k$ an integer. The product and quotient laws do the arithmetic: $(6 \times 10^{8})(5 \times 10^{-3}) = 30 \times 10^{5} = 3 \times 10^{6}$.
Formulas and Theorems
- Laws for rational exponents ($a, b > 0$): $a^m a^n = a^{m+n},\quad \dfrac{a^m}{a^n} = a^{m-n},\quad (a^m)^n = a^{mn}$
- Power of a product or quotient: $(ab)^n = a^n b^n,\quad \left(\dfrac ab\right)^{n} = \dfrac{a^n}{b^n},\quad \left(\dfrac ab\right)^{-n} = \left(\dfrac ba\right)^{n}$
- Roots as exponents: $a^{1/n} = \sqrt[n]{a},\quad a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$
Worked Examples and Your Turn
Study each example, then try the paired problem.
Example 1: A power of a product
Problem: Simplify $(-2x^3y^2)^3 \cdot 4xy^{-5}$ so that no exponent is negative.
Solution:
- The cube reaches every factor inside the parentheses: $(-2)^3 = -8$, $(x^3)^3 = x^9$, $(y^2)^3 = y^6$. So the first part is $-8x^9y^6$. Why: $(ab)^n = a^nb^n$.
- Multiply the numbers: $-8 \cdot 4 = -32$.
- Add exponents on each letter: $x^{9+1} = x^{10}$ and $y^{6+(-5)} = y^1$.
Answer: $-32x^{10}y$
> Notice: The number in front is a factor like any other. The most common slip is to cube the letters and forget the $-2$.
> Verification & Check: Put $x = 1$, $y = 2$ into the original: $(-2 \cdot 4)^3 \cdot 4 \cdot 2^{-5} = -512 \cdot \frac{4}{32} = -64$. The answer gives $-32 \cdot 1 \cdot 2 = -64$. Same.
Your Turn 1: Simplify $(3a^2b^{-1})^2 \cdot 2a^{-3}b^4$ with positive exponents.
Example 2: A negative power of a quotient
Problem: Simplify $\left(\dfrac{2x^{-2}y^3}{x^4y}\right)^{-3}$ so that no exponent is negative.
Solution:
- Tidy the inside first. For $x$: $-2 - 4 = -6$. For $y$: $3 - 1 = 2$. The inside is $2x^{-6}y^2$.
- Raise every factor to the $-3$: $2^{-3}\,x^{(-6)(-3)}\,y^{2(-3)} = 2^{-3}x^{18}y^{-6}$. Why: a power of a power multiplies the exponents, signs included.
- Send the negative exponents below the line: $\dfrac{x^{18}}{8y^6}$.
Answer: $\dfrac{x^{18}}{8y^6}$
> Verification & Check: Put $x = 1$, $y = 2$: the inside is $\frac{2 \cdot 8}{2} = 8$, and $8^{-3} = \frac{1}{512}$. The answer gives $\frac{1}{8 \cdot 64} = \frac{1}{512}$. Same.
Your Turn 2: Simplify $\left(\dfrac{3a^{2}}{a^{-1}b^{2}}\right)^{-2}$ with positive exponents.
Example 3: Rational exponents by hand
Problem: Evaluate $32^{4/5}$, $125^{-2/3}$ and $\left(\dfrac{16}{81}\right)^{3/4}$ without a calculator.
Solution:
- $32^{4/5}$: the $5$ is a fifth root, $\sqrt[5]{32} = 2$. Then the power: $2^4 = 16$.
- $125^{-2/3}$: the minus sign makes a reciprocal, so find $125^{2/3}$ first. $\sqrt[3]{125} = 5$ and $5^2 = 25$. The value is $\frac{1}{25}$.
- $\left(\frac{16}{81}\right)^{3/4}$: the root and power reach top and bottom. $\sqrt[4]{16} = 2$ and $\sqrt[4]{81} = 3$, so the fourth root is $\frac23$, and $\left(\frac23\right)^3 = \frac{8}{27}$.
Answer: $16$; $\dfrac{1}{25}$; $\dfrac{8}{27}$
> Notice: Take the root before the power. $32^4 = 1{,}048{,}576$ is a miserable number to take a fifth root of; $2^4$ is not.
Your Turn 3: Evaluate $64^{5/6}$ and $49^{-3/2}$.
Example 4: Radicals with variables as powers
Problem: For $x > 0$, write $\sqrt[3]{x^2} \cdot \sqrt{x}$ as a single power of $x$, and then as one radical. Then simplify $\left(8x^6\right)^{2/3}$.
Solution:
- Change each radical to a power: $\sqrt[3]{x^2} = x^{2/3}$ and $\sqrt{x} = x^{1/2}$.
- Multiply by adding the exponents over a common denominator: $\frac23 + \frac12 = \frac46 + \frac36 = \frac76$. So $x^{7/6}$.
- Back to a radical: the denominator $6$ is the root, the numerator $7$ is the power. $x^{7/6} = \sqrt[6]{x^7}$.
- For $\left(8x^6\right)^{2/3}$, the power reaches both factors: $8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 4$ and $\left(x^6\right)^{2/3} = x^{4}$.
Answer: $x^{7/6} = \sqrt[6]{x^7}$; $4x^4$
> Verification & Check: Try $x = 64$: $\sqrt[3]{4096} = 16$ and $\sqrt{64} = 8$, product $128$. And $64^{7/6} = \left(\sqrt[6]{64}\right)^7 = 2^7 = 128$. Same.
Your Turn 4: For $x > 0$, write $\dfrac{\sqrt[4]{x^3}}{\sqrt{x}}$ as a single power of $x$. Then simplify $\left(27y^9\right)^{1/3}$.
Example 5: Three slips with powers and roots
Problem: Each line has an error. Find it and fix it: $(3x^2)^3 = 3x^6$; $16^{-1/2} = -4$; $(x^2 + 9)^{1/2} = x + 3$.
Solution:
- Common mistake: the outside exponent is applied to the letters only, the minus in the exponent is read as a negative answer, and a root is split across a plus sign.
- Do this instead. $(3x^2)^3 = 3^3(x^2)^3 = 27x^6$. Why: the cube reaches the $3$ as well.
- $16^{-1/2} = \dfrac{1}{16^{1/2}} = \dfrac{1}{4}$. Why: a negative exponent makes a reciprocal, never a negative number.
- $(x^2 + 9)^{1/2} = \sqrt{x^2 + 9}$, and it does not simplify. Test $x = 4$: $\sqrt{25} = 5$, but $x + 3 = 7$. Why: a power spreads over a product, not over a sum.
Answer: $27x^6$; $\dfrac14$; $\sqrt{x^2 + 9}$ stays as it is
> Notice: One number put into the claimed identity is enough to kill it. Keep $x = 4$ or $x = 2$ ready for this.
Your Turn 5: Fix the errors: $(-2y^3)^2 = -4y^6$ and $8^{-2/3} = -4$.
Example 6: Solve by a common base
Problem: Solve $4^{x+1} = 8^{x}$ and $9^{x} = \dfrac{1}{27}$.
Solution:
- Write both sides of the first as powers of $2$: $4^{x+1} = \left(2^2\right)^{x+1} = 2^{2x+2}$ and $8^x = 2^{3x}$.
- Equal powers of the same base have equal exponents: $2x + 2 = 3x$, so $x = 2$.
- For the second use base $3$: $9^x = 3^{2x}$ and $\frac{1}{27} = 3^{-3}$. So $2x = -3$ and $x = -\frac32$.
Answer: $x = 2$; $x = -\dfrac32$
> Verification & Check: $4^3 = 64$ and $8^2 = 64$. And $9^{-3/2} = \dfrac{1}{\left(\sqrt9\right)^3} = \dfrac{1}{27}$. Both check.
Your Turn 6: Solve $25^{x} = 125^{x-1}$ and $32^{x} = \dfrac14$.
Example 7: A fraction of a doubling time
Problem: A culture starts with $500$ bacteria and doubles every $3$ hours, so $N = 500 \cdot 2^{t/3}$ after $t$ hours. (a) Find $N$ after $12$ hours. (b) Write $N$ after $1$ hour using a radical, and estimate it. (c) By what factor does the culture grow each hour?
Solution:
- (a) $t = 12$ gives $2^{12/3} = 2^4 = 16$, so $N = 500 \cdot 16 = 8000$.
- (b) $t = 1$ gives $N = 500 \cdot 2^{1/3} = 500\sqrt[3]{2}$. Since $\sqrt[3]{2} \approx 1.26$, $N \approx 630$.
- (c) Each hour adds $1$ to $t$, so the power of $2$ goes up by $\frac13$: the culture is multiplied by $2^{1/3} \approx 1.26$ every hour. Why: $2^{(t+1)/3} = 2^{t/3} \cdot 2^{1/3}$.
Answer: (a) $8000$; (b) $500\sqrt[3]{2} \approx 630$; (c) about $1.26$, a $26\%$ increase per hour
> Notice: The hourly factor is not $\frac23$, and it is not a third of $2$. It is the number that, used as a factor three times, makes $2$.
> Verification & Check: Three hours of the hourly factor: $\left(2^{1/3}\right)^3 = 2$. One doubling, as promised.
Your Turn 7: A town's population is $P = 1000 \cdot 3^{t/2}$ after $t$ years. Find $P$ at $t = 4$ and the yearly growth factor.
The chapter continues in the book.
