QuanticaPrealgebraOpen in the courseBack in section 2.2 you found that raising a power to a power multiplies the two exponents. That one rule stretches much further than it first looks. It still works when three or more exponents are stacked, when the inside is a fraction, when an exponent turns negative or zero, and even when a number sits out in front of the variable. This last section puts every law in the chapter to work at once.
Problem
A glassblower mixes a colour by the ratio \(\dfrac{7}{2}\), and a deeper batch needs that ratio cubed, \(\left(\dfrac{7}{2}\right)^3\). Find it two ways. First multiply the fraction by itself three times. Then cube the top and the bottom on their own, \(\dfrac{7^3}{2^3}\). Give the value as a fraction.
Show a hint
- Multiplying \(\dfrac{7}{2}\times\dfrac{7}{2}\times\dfrac{7}{2}\) collects every top together and every bottom together.
- The tops give \(7^3=343\) and the bottoms give \(2^3=8\). Both methods build the same fraction.
Show the full solution
Multiplying the fraction by itself three times gives three 7s on top and three 2s on the bottom. $$\left(\dfrac{7}{2}\right)^3 = \dfrac{7\times 7\times 7}{2\times 2\times 2} = \dfrac{7^3}{2^3} = \dfrac{343}{8}$$ The value is \(\boxed{\dfrac{343}{8}}\). Cubing the top and the bottom on their own is the same work written more directly.
Problem
Test the same idea on a fourth power so it is not just a quirk of cubing. A kite design scales the ratio \(\dfrac{3}{5}\) to the fourth power. Send the exponent onto the top and the bottom separately, \(\dfrac{3^4}{5^4}\), and give the value as a fraction.
Show a hint
- The exponent lands on both the top and the bottom, so work out \(3^4\) and \(5^4\) each.
- One of them is \(3^4 = 81\). Work out \(5^4\) yourself and stack it underneath.
Show the full solution
Send the exponent onto the top and the bottom separately. $$\left(\frac{3}{5}\right)^4 = \frac{3^4}{5^4} = \frac{81}{625}$$ The value is \(\boxed{\dfrac{81}{625}}\). Four copies of \(\frac{3}{5}\) gather four 3s on top and four 5s on the bottom, so the rule is not special to cubing.
Two stacked exponents multiply. Three stacked exponents do the same thing, since the rule just fires twice. Peel the outermost exponent off first, then handle what is left.
Problem
A folding screen carries a motif repeated \(2^2\) times across one panel. A room divider holds \((2^2)^3\) such panels in a grid, and an exhibition lines up \(\big((2^2)^3\big)^2\) of those dividers. Simplify the whole tower \(\big((2^2)^3\big)^2\) to a single power of 2 and give the exponent.
Show a hint
- Peel off the outermost exponent first. The power rule turns \(\big((2^2)^3\big)^2\) into \((2^2)^{3\times 2}\).
- Now \((2^2)^6 = 2^{2\times 6}\). All three exponents end up multiplied together.
Show the full solution
Work from the outside in, using the power rule each time. $$\big((2^2)^3\big)^2 = (2^2)^{3\times 2} = (2^2)^6 = 2^{2\times 6} = 2^{12}$$ The exponent is \(\boxed{12}\). Every exponent in the tower ends up multiplied together, \(2\times 3\times 2 = 12\).
Problem
An elevator panel shows the reading \((9^4)^{0}\). Without a calculator, give its exact value.
Show a hint
- Multiply the stacked exponents first, \(4 \times 0\), or recall what any nonzero base to the zero power is.
- Whichever route you take, the panel settles on the value that every nonzero base shares at the zero power.
Show the full solution
Multiply the exponents, \((9^4)^0 = 9^{4\times 0} = 9^0 = 1\), so the panel reads \(\boxed{1}\). You can also skip the multiplying, since any nonzero base to the zero power is already 1.
Watch your step here, because this is the single most common slip in the whole chapter. Stacking exponents multiplies them, while multiplying powers of the same base adds them. They look alike on the page and do opposite things to the exponents.
Problem
A cargo line runs two belts. The upper belt batches packages as \((5^3)^2\), and the lower belt batches them as \(5^3 \times 5^2\). They are not the same count. Rewrite each as a single power of 5, then give the difference of the two exponents, the upper belt exponent minus the lower belt exponent.
Show a hint
- The upper belt is a power raised to a power, so multiply its exponents. The lower belt is a product of powers, so add its exponents.
- Upper is \(5^{3\times 2}=5^6\) and lower is \(5^{3+2}=5^5\). Subtract the two exponents.
Show the full solution
The upper belt stacks a power on a power, so the exponents multiply, \((5^3)^2 = 5^{3\times 2} = 5^6\). The lower belt multiplies two powers of the same base, so the exponents add, \(5^3\times 5^2 = 5^{3+2} = 5^5\). The difference is \(6-5 = \boxed{1}\). These are the two rules people mix up most, and they do opposite things to the exponents.
The power rule never asked the inner exponent to be positive. Try it with a negative inner exponent and check that multiplying the exponents still gives the right answer.
Problem
A sound desk stores the faint gain \(2^{-3}\), then a sharpening step squares whatever is stored, computing \((2^{-3})^2\). Multiply the exponents just as always and write the result as a single power of 2 with one exponent. Give that exponent.
Show a hint
- The power rule multiplies the exponents even when one of them is negative.
- Carry the sign through the multiplication, so the inner \(-3\) meets the outer \(2\). The product is the new exponent.
Show the full solution
Multiply the exponents, sign and all. $$(2^{-3})^2 = 2^{(-3)\times 2} = 2^{-6}$$ The exponent is \(\boxed{-6}\). Written without a negative that is \(\dfrac{1}{2^6} = \dfrac{1}{64}\), so squaring a small number made it smaller still.
Now put a number in front of the variable. When an outer exponent sits on something like \(2x^3\), it is tempting to leave the \(2\) alone. You cannot. Everything inside the parentheses is a product, and the coefficient is one of its factors, so the exponent lands on it too.
Problem
A bakery stamps trays of cookies, each tray holding \(2x^3\) rows where \(x\) is the dough setting. A holiday order raises that whole tray to the fourth power, \((2x^3)^4\). One apprentice writes \(2x^{12}\) and another writes \(16x^{12}\). Exactly one is right. Give the correct simplified form. Type it in the form such as 16x^12.
Show a hint
- The outer 4 grips every factor inside the parentheses, both the \(2\) and the \(x^3\). Do not leave the \(2\) untouched.
- One factor gives \(2^4 = 16\). Send the same \(4\) onto \(x^3\) yourself and combine the two pieces.
Show the full solution
The outer exponent lands on every factor inside, the coefficient included. $$(2x^3)^4 = 2^4 \times (x^3)^4 = 16 \times x^{12} = 16x^{12}$$ The correct form is \(\boxed{16x^{12}}\). The apprentice who wrote \(2x^{12}\) forgot to raise the \(2\).
One more case ties the negative powers from the last section to the quotient rule from this one. Make the outer exponent negative while it sits on a fraction, and the fraction flips over.
Problem
A recipe scales to \(\dfrac{2}{3}\) strength, but a reset raises that whole setting to the power \(-3\). Evaluate \(\left(\dfrac{2}{3}\right)^{-3}\). Treat the negative exponent as a reciprocal first, flipping \(\dfrac{2}{3}\) over, then cube. Give the value as a fraction.
Show a hint
- A negative exponent takes the reciprocal of the base, and the reciprocal of \(\dfrac{2}{3}\) is \(\dfrac{3}{2}\).
- So \(\left(\dfrac{2}{3}\right)^{-3} = \left(\dfrac{3}{2}\right)^3 = \dfrac{3^3}{2^3}\).
Show the full solution
The negative exponent flips the fraction and turns positive. $$\left(\dfrac{2}{3}\right)^{-3} = \left(\dfrac{3}{2}\right)^{3} = \dfrac{3^3}{2^3} = \dfrac{27}{8}$$ The answer is \(\boxed{\dfrac{27}{8}}\). Flip first, then send the exponent onto the new top and the new bottom.
These laws can all show up in a single expression. A power of a quotient, a raised coefficient, and the quotient rule can sit in the same problem, with order of operations deciding what comes first. Work one grouped piece at a time.
Problem
A lighting rig simplifies \(\dfrac{(3x^4)^2}{x^5}\) to a single coefficient and power of \(x\). Find it. Type it in the form such as 16x^12.
Show a hint
- Raise the numerator first. The outer 2 lands on the \(3\) and on the \(x^4\).
- The top becomes \(9x^8\). Now divide by \(x^5\) with the quotient rule, subtracting the exponents, and keep the \(9\).
Show the full solution
Open the numerator, letting the outer exponent hit both factors. $$(3x^4)^2 = 3^2 \times x^{4\times 2} = 9x^8$$ Now divide by \(x^5\). The coefficient is untouched by the division, and the quotient rule subtracts the variable exponents. $$\dfrac{9x^8}{x^5} = 9x^{8-5} = 9x^3$$ The answer is \(\boxed{9x^3}\).
Problem
A power grid balances two terms. Simplify \(\left(\dfrac{x^3}{2}\right)^{2} \times \left(\dfrac{x^2}{2}\right)^{-3}\) to a single number with no variable left. Find that number.
Show a hint
- The first term lands on the top and the bottom, \(\dfrac{(x^3)^2}{2^2} = \dfrac{x^6}{4}\). The second has a negative outer exponent, so flip it first.
- Flipping gives \(\left(\dfrac{2}{x^2}\right)^{3} = \dfrac{8}{x^6}\). Now multiply the two fractions and watch the \(x\) cancel.
Show the full solution
The first term sends the exponent onto the top and the bottom. $$\left(\dfrac{x^3}{2}\right)^{2} = \dfrac{(x^3)^2}{2^2} = \dfrac{x^6}{4}$$ The second has a negative outer exponent, so flip the fraction first, then raise it, coefficient and all. $$\left(\dfrac{x^2}{2}\right)^{-3} = \left(\dfrac{2}{x^2}\right)^{3} = \dfrac{2^3}{(x^2)^3} = \dfrac{8}{x^6}$$ Multiply the two results. $$\dfrac{x^6}{4} \times \dfrac{8}{x^6} = \dfrac{8x^6}{4x^6} = 2x^{6-6} = 2$$ The answer is \(\boxed{2}\).
Here is the capstone, one expression that uses every law in the chapter. There is a coefficient inside a flip, a quotient under an exponent, a tower of three powers, and the quotient rule. Work one grouped piece at a time.
Problem
Evaluate $$\left(\dfrac{(2x^3)^2}{4x^4}\right)^{3} \times \left(\dfrac{x^2}{2}\right)^{-3}.$$ It is built so the variable cancels completely and a single number remains. Find that number.
Show a hint
- Simplify deep inside the first parentheses before applying the outer 3. Raise \((2x^3)^2 = 4x^6\), then divide by \(4x^4\) to get a clean \(x^2\) with coefficient \(1\).
- The first factor becomes \((x^2)^3 = x^6\). The second has a negative outer exponent, so flip \(\dfrac{x^2}{2}\) to \(\dfrac{2}{x^2}\) before cubing. Then multiply and watch the \(x\) disappear.
Show the full solution
Work the first factor from the inside out. The inner top is \((2x^3)^2 = 2^2 x^6 = 4x^6\), and dividing by \(4x^4\) gives $$\dfrac{4x^6}{4x^4} = \dfrac{4}{4}\,x^{6-4} = x^2.$$ The outer exponent then multiplies, \(\left(x^2\right)^{3} = x^{6}\). The second factor has a negative outer exponent, so flip the fraction, then raise top and bottom. $$\left(\dfrac{x^2}{2}\right)^{-3} = \left(\dfrac{2}{x^2}\right)^{3} = \dfrac{2^3}{(x^2)^3} = \dfrac{8}{x^6}$$ Multiply the two pieces. $$x^6 \times \dfrac{8}{x^6} = 8\,x^{6-6} = 8x^0 = 8$$ The number is \(\boxed{8}\).
Practice these ideas
Practice
You already know that stacking a power on a power multiplies the two exponents. This section pushes that idea everywhere it can go, onto fractions, onto coefficients, up nested towers, and across negative exponents, until every law in the chapter can show up in one expression at once.
Practice
A canyon guide tests an echo by stacking it on itself. One shout returns as \(\left(4^2\right)^3\) faint repeats. Rewrite that as a single power of \(4\) and give the exponent.
Show the solution
Stacking a power on a power multiplies the two exponents rather than adding them. $$\left(4^2\right)^3 = 4^{2\times 3} = 4^6$$ The exponent is \(\boxed{6}\). The outer \(3\) asks for three copies of \(4^2\), and each copy carries two factors of \(4\), so there are \(2\times 3=6\) factors in all.
Practice
A quilter sews a block by raising a small square to a power. The block holds \(\left(\dfrac{2}{3}\right)^4\) of one unit of fabric. Write that as a single fraction.
Show the solution
An outer exponent on a quotient lands on both the top and the bottom. $$\left(\dfrac{2}{3}\right)^4 = \dfrac{2^4}{3^4} = \dfrac{16}{81}$$ The block is \(\boxed{\dfrac{16}{81}}\) of a unit. Four copies of \(\dfrac{2}{3}\) multiplied together gather four \(2\)s on top and four \(3\)s on the bottom.
Practice
An outer exponent on a fraction lands on the top and the bottom alike, so \(\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\) for any nonzero \(b\). It works because multiplying \(n\) copies of \(\dfrac{a}{b}\) gathers \(n\) factors of \(a\) on top and \(n\) factors of \(b\) on the bottom.
Practice
On a beadwork loom a single tile pattern is enlarged by raising it to the fourth power. The tile is described by \(2b^3\), so the enlarged pattern is \(\left(2b^3\right)^4\). Simplify it to a coefficient times a single power of \(b\).
Show the solution
The outer exponent reaches every factor inside, coefficient included. $$\left(2b^3\right)^4 = 2^4 \times \left(b^3\right)^4 = 16 \times b^{3\times 4} = 16b^{12}$$ The pattern simplifies to \(\boxed{16b^{12}}\). It is not \(2b^{12}\), and forgetting to raise the \(2\) is the most common slip here.
Practice
When a product inside parentheses carries a number out front, the outer exponent lands on that number too, so \(\left(k\,x^m\right)^n=k^n\,x^{mn}\). For example \(\left(2x^3\right)^4=16x^{12}\), never \(2x^{12}\). The coefficient is a factor like any other, so it cannot be left behind.
Practice
Two coral reef readings look alike but mean different things. The first colony grows as \(\left(5^2\right)^3\) polyps, a power stacked on a power. The second grows as \(5^2 \times 5^3\) polyps, two powers multiplied together. Both fold to a single power of \(5\). By how much does the first reading's exponent beat the second reading's exponent?
Show the solution
Stacking a power on a power multiplies the exponents, while multiplying two powers of the same base adds them. $$\left(5^2\right)^3 = 5^{2\times 3} = 5^6, \qquad 5^2 \times 5^3 = 5^{2+3} = 5^5$$ The first reading reaches exponent \(6\) and the second reaches \(5\), so stacking pulls ahead by \(6-5=\boxed{1}\).
Practice
A lantern festival dims a light by raising a faint setting to a power. The setting is \(2^{-3}\) of full brightness, and squaring it gives \(\left(2^{-3}\right)^2\). Write the result as a single power of \(2\) and give the exponent.
Show the solution
Multiply the two exponents, sign and all. $$\left(2^{-3}\right)^2 = 2^{(-3)\times 2} = 2^{-6}$$ The exponent is \(\boxed{-6}\). Two copies of \(2^{-3}\) put six factors of \(2\) in the denominator, which is \(\dfrac{1}{64}\).
Practice
Stacking really does outrun adding. Starting from \(10^4\), multiplying by another \(10^3\) only reaches \(10^7\), but raising it to the \(3\)rd power leaps to \(10^{12}\). That gap of five zeros is the difference between joining copies and making copies of copies.
Practice
A model railway sets its scale as a fraction, then a reversing switch raises that scale to the power \(-2\). The scale is \(\dfrac{3}{5}\), so the switch produces \(\left(\dfrac{3}{5}\right)^{-2}\). Write the result as a single fraction.
Show the solution
A negative outer exponent flips the fraction, then the exponent goes positive. $$\left(\dfrac{3}{5}\right)^{-2} = \left(\dfrac{5}{3}\right)^{2} = \dfrac{5^2}{3^2} = \dfrac{25}{9}$$ The switch produces \(\boxed{\dfrac{25}{9}}\). The flip handles the minus sign, and the squaring lands on the new top and the new bottom.
Practice
A negative outer exponent on a fraction turns it upside down and makes the exponent positive, so \(\left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^{n}\) for nonzero \(a\) and \(b\). The minus sign is really just the reciprocal rule applied to a fraction.
Practice
A fractal antenna is etched in stages. Each stage splits every arm into \(2^2\) smaller arms, and the design runs through three stages, so the tip count builds up as the nested tower \(\left(\left(2^2\right)^2\right)^2\). When three exponents are stacked, all of them multiply together. Rewrite the tower as a single power of \(2\) and give its exponent.
Show the solution
Every stacked exponent multiplies into the next. $$\left(\left(2^2\right)^2\right)^2=2^{\,2\times2\times2}=2^{8}$$ The exponent is \(\boxed{8}\), so the tip count is \(256\).
Practice
A salt cavern records its volume as \(\left(2c^2\right)^3\) and the brine it holds as \(4c^5\). The packing density is the volume divided by the brine, \(\dfrac{\left(2c^2\right)^3}{4c^5}\). Simplify it to a coefficient times a single power of \(c\).
Show the solution
Open the top with the power of a product rule, raising the coefficient too. $$\left(2c^2\right)^3 = 2^3 \times c^{2\times 3} = 8c^6$$ Now divide, splitting the coefficients from the powers of \(c\). $$\dfrac{8c^6}{4c^5} = \dfrac{8}{4} \times c^{6-5} = 2c^1 = 2c$$ The packing density simplifies to \(\boxed{2c}\).
Practice
A tuning fork's overtone strength is modeled by \(\left(a^2\right)^3\times\left(a^{-3}\right)^2\) for a nonzero \(a\). Simplify each tower first, then combine. The messy expression collapses to something tiny. What is its value?
Show the solution
Multiply within each tower, then add when the powers meet. $$\left(a^2\right)^3=a^{6},\qquad \left(a^{-3}\right)^2=a^{-6}$$ $$a^{6}\times a^{-6}=a^{6+(-6)}=a^{0}=1$$ Any nonzero base to the zero power is \(1\), so the whole thing folds to \(\boxed{1}\).
Practice
A tower of exponents grows almost too fast to picture. Stacking just \(\left(\left(\left(2^2\right)^2\right)^2\right)^2\) multiplies four twos to \(2^{16}\), already past sixty thousand. A few more floors and the number would dwarf the count of grains of sand on Earth.
Practice
A wind chime nests three tunings. The deepest tube rings as \(\left(\left(2^2\right)^3\right)^2\) and a damper divides that ring by \(\left(2^4\right)^2\). The audible tone is \(\dfrac{\left(\left(2^2\right)^3\right)^2}{\left(2^4\right)^2}\). Write it as a single power of \(2\) and give its value.
Show the solution
Collapse each tower by multiplying all of its stacked exponents. On top, three exponents multiply. $$\left(\left(2^2\right)^3\right)^2 = 2^{2\times 3\times 2} = 2^{12}$$ On the bottom, two multiply. $$\left(2^4\right)^2 = 2^{4\times 2} = 2^{8}$$ Now divide a common base by subtracting. $$\dfrac{2^{12}}{2^{8}} = 2^{12-8} = 2^{4} = 16$$ The audible tone is \(\boxed{16}\).
Practice
A foundry mixes powers of \(2\) and \(5\) into one casting reading, $$\left(\dfrac{2^3 \times 5}{10^2}\right)^{-2}.$$ Simplify the whole thing to a single fraction.
Show the solution
Settle the inside first. The top is \(2^3 \times 5 = 8\times 5 = 40\) and the bottom is \(10^2 = 100\). $$\dfrac{2^3 \times 5}{10^2} = \dfrac{40}{100} = \dfrac{2}{5}$$ Now the negative outer exponent flips the fraction, then squares it on both top and bottom. $$\left(\dfrac{2}{5}\right)^{-2} = \left(\dfrac{5}{2}\right)^{2} = \dfrac{25}{4}$$ The casting reading is \(\boxed{\dfrac{25}{4}}\).
Practice
A beehive arranges its active brood cells by a single formula, $$\dfrac{\left(2^3\right)^4\times\left(2^{-2}\right)^3}{\left(2^4\right)^2}.$$ Every law in this chapter shows up once. Collapse it all into a single power of \(2\) and give that power's exponent.
Show the solution
Flatten every stacked power first, multiplying the exponents and carrying the signs. $$\left(2^3\right)^4=2^{12},\qquad \left(2^{-2}\right)^3=2^{-6},\qquad \left(2^4\right)^2=2^{8}$$ Multiply the two powers on top by adding their exponents. $$2^{12}\times2^{-6}=2^{12+(-6)}=2^{6}$$ Then divide by subtracting the bottom exponent. $$\dfrac{2^{6}}{2^{8}}=2^{6-8}=2^{-2}$$ The exponent is \(\boxed{-2}\), and the whole formula is \(\dfrac{1}{4}\).