QuanticaPrealgebraOpen in the courseYou can always compute \(2^{13}\) by multiplying thirteen 2s in a row. But for \(2^{100}\) that approach is hopeless. Luckily the exponents themselves follow a few simple rules that let you combine powers without ever writing out the factors.
Problem
A strip of paper starts as a single layer. You fold it in half 3 times, creating \(2^3\) layers. Without unfolding, you fold it 4 more times. How many layers does it have now?
Show a hint
- After 3 folds you have \(2^3 = 8\) layers. Each additional fold doubles every existing layer.
Show the full solution
After 3 folds the stack is \(2^3 = 8\) layers, and each of the 4 new folds doubles every layer, multiplying the stack by \(2^4 = 16\). $$2^3 \times 2^4 = 8 \times 16 = \boxed{128}$$ Three doublings followed by four more is seven doublings in a row, which is exactly why the exponents add.
Problem
Find the missing exponent. $$6^{\,?} \times 6^8 = 6^{15}$$
Show a hint
- The two exponents must add to 15.
Show the full solution
The two exponents on the left have to add to the exponent on the right, so \(? + 8 = 15\) and \(? = 15 - 8 = \boxed{7}\). The product rule runs both ways. Forward it merges two powers into one, backward it recovers a missing exponent.
The product rule handles multiplication. Division works the other way. Instead of collecting factors you cancel them, and the exponents get subtracted. The next two problems show why.
Problem
Verify the subtraction pattern by cancelling factors directly. Write \(\dfrac{5^7}{5^4}\) as a fraction with every factor of 5 listed in numerator and denominator, then cancel matching pairs. The survivor is a single power of 5. What is its value?
Show a hint
- List seven 5s on top and four 5s on the bottom, then cancel matching pairs.
Show the full solution
Write out every factor, seven 5s on top and four on the bottom, then cancel. $$\frac{5^7}{5^4} = \frac{5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5}{5 \cdot 5 \cdot 5 \cdot 5} = 5^3 = \boxed{125}$$ Four cancellations remove four factors from the top and three 5s survive. Subtracting exponents is just counting the survivors.
Problem
Simplify \(\dfrac{2^{13} \times 2^4}{2^{10}}\) to a single power of 2, then give its value.
Show a hint
- Combine the numerator with the product rule first, then apply the quotient rule.
Show the full solution
Combine the numerator first, then divide. $$\frac{2^{13} \times 2^4}{2^{10}} = \frac{2^{17}}{2^{10}} = 2^{17-10} = 2^7 = \boxed{128}$$ Multiplying same-base powers adds the exponents and dividing them subtracts, so the two laws never get in each other's way.
The product and quotient rules both combine two separate powers. A power can also be raised to another power, like \((3^4)^5\). That case has a rule of its own, and the next problem works it out.
Problem
Compute \((2^3)^4\) by first evaluating \(2^3\), then raising that result to the 4th power. You should arrive at 4096. Write 4096 as a power of 2 and look at the relationship between that final exponent and the original two exponents, 3 and 4.
Show a hint
- \(2^3 = 8\). Now compute \(8^4\).
- \(4096 = 2^{12}\). How does 12 relate to 3 and 4?
Show the full solution
\(2^3 = 8\), and \(8^4 = \boxed{4096}\). Written as a power of 2 that is \(4096 = 2^{12}\), and \(12 = 3 \times 4\). The outer exponent asks for four copies of \(2^3\), each copy brings 3 factors of 2, so there are 12 factors in all. Stacked exponents multiply.
Problem
Since \(64 = 2^6\), rewrite \(64^{10}\) as a single power of 2. What is the exponent?
Show a hint
- Substitute \(64 = 2^6\), then apply the power rule.
Show the full solution
Since \(64 = 2^6\), $$64^{10} = (2^6)^{10} = 2^{6 \times 10} = 2^{60},$$ so the exponent is \(\boxed{60}\). Rewriting an unfamiliar base as a power of a known base and then multiplying the exponents is the standard move when both numbers are built from the same one.
Problem
A square floor is tiled with square tiles. Each tile has a side length of \(4^2\) centimetres, and the floor has \(4^3\) tiles along each side. Find the area of the floor in square centimetres, expressed as \(4^n\). What is \(n\)?
Show a hint
- The floor's side length in centimetres is \(4^3\) tiles each \(4^2\) cm wide. Find the total side length in cm first.
- Area equals side length squared. Apply the power rule.
Show the full solution
One side of the floor holds \(4^3\) tiles, each \(4^2\) cm wide. $$4^3 \times 4^2 = 4^{3+2} = 4^5 \text{ cm per side}$$ The floor is square, so the area is that side length squared. $$(4^5)^2 = 4^{5 \times 2} = 4^{10} \text{ sq cm}$$ So \(n = \boxed{10}\). Product rule to build the side, power rule to square it.
The three rules so far all require the same base throughout. But a power can also sit over a product, like \((2 \times 5)^3\). The next problem asks whether the exponent distributes over the multiplication or not.
Problem
Compute \((2 \times 5)^3\) two ways: first as \(10^3\), then as \(2^3 \times 5^3\). Do you get the same answer?
Show a hint
- \(10^3 = 1000\). Now find \(2^3 \times 5^3 = 8 \times 125\).
Show the full solution
Collapsing the base first, \(2 \times 5 = 10\) and \(10^3 = 1000\). Distributing the exponent instead, \(2^3 \times 5^3 = 8 \times 125 = 1000\). Both routes give \(\boxed{1000}\). Cubing the pair and multiplying the cubes count the same factors, and that is the power-of-a-product rule.
Problem
Find the positive integer \(n\) such that \((9^n)^4 = 9^{20}\).
Show a hint
- The power rule gives \((9^n)^4 = 9^{n \times 4}\). What must \(n \times 4\) equal?
Show the full solution
The power rule multiplies the stacked exponents, so \((9^n)^4 = 9^{n \times 4}\). For that to equal \(9^{20}\) the exponents must match, giving \(n \times 4 = 20\) and \(n = \boxed{5}\). Backward, the same rule divides. Exponent 20 outside a 4th power means the inner exponent was \(20 \div 4\).
Problem
Compute \(6^4 \div (2^4 \times 3^4)\) without multiplying anything out.
Show a hint
- Write \(6 = 2 \times 3\) and apply the power-of-a-product rule to \(6^4\).
Show the full solution
Since \(6 = 2 \times 3\), the outer exponent drops onto each factor. $$6^4 = (2 \times 3)^4 = 2^4 \times 3^4$$ The numerator is now identical to the denominator, so the quotient is \(\boxed{1}\). No multiplying out was needed once \(6^4\) was seen to already contain those factors.
You now have four tools: add exponents when multiplying same-base powers, subtract when dividing, multiply when raising a power to a power, and distribute when a power sits over a product. The last two problems combine all four.
Problem
Simplify \(\dfrac{(3^5)^4}{3^{11} \times 3^5}\) to a single power of 3, then give its value.
Show a hint
- Apply the power rule to the numerator and the product rule to the denominator. Then divide.
Show the full solution
The numerator is \((3^5)^4 = 3^{5 \times 4} = 3^{20}\) and the denominator is \(3^{11} \times 3^5 = 3^{16}\). $$\frac{3^{20}}{3^{16}} = 3^{20-16} = 3^4 = \boxed{81}$$ Power rule on top, product rule on the bottom, quotient rule to finish. Deciding which law belongs to which piece before doing any arithmetic is the whole skill.
Problem
Compute $$\frac{6^{20} \times 4^{15}}{8^{10} \times 3^{20}}.$$ The result is a power of 2. What is the exponent?
Show a hint
- Write every base as a product of primes: \(6 = 2 \times 3\), \(4 = 2^2\), \(8 = 2^3\). Apply the power-of-a-product rule to each.
- After substituting, separate the powers of 2 from the powers of 3.
- The powers of 3 in the numerator and denominator are equal and cancel completely.
Show the full solution
Rewrite every base in primes. \(6^{20} = (2 \times 3)^{20} = 2^{20} \times 3^{20}\), \(4^{15} = (2^2)^{15} = 2^{30}\), and \(8^{10} = (2^3)^{10} = 2^{30}\). $$\frac{2^{20} \times 3^{20} \times 2^{30}}{2^{30} \times 3^{20}} = \frac{2^{50} \times 3^{20}}{2^{30} \times 3^{20}} = 2^{50-30} = 2^{20}$$ The exponent is \(\boxed{20}\). The 3s matched top and bottom and cancelled. Break every base into primes first and the laws handle the rest.
Practice these ideas
Practice
A kaleidoscope adds mirrors in two stages. The first tube holds \(4\) mirrors, and each mirror doubles the number of visible copies of a bead, so the tube shows \(2^4\) copies. You twist on a second tube of \(6\) more doubling mirrors. Written as a single power of \(2\), how many copies of the bead do you see through the whole instrument?
Show the solution
Each mirror doubles, so the first tube multiplies by \(2^4\) and the second by \(2^6\). Sending one bead through both means doubling \(4\) times, then \(6\) more, which is \(10\) doublings all told. Multiplying powers of the same base adds the exponents. $$2^4\times2^6=2^{4+6}=\boxed{2^{10}}$$ That is \(1024\) copies, a number you never had to build by hand.
Practice
That \(2^{10}=1024\) is the reason a kilobyte is not quite \(1000\) bytes. Computers count in doublings, and ten doublings overshoot a thousand by just \(24\). Engineers shrugged, borrowed the word kilo anyway, and the small fib has stuck ever since.
Practice
A library's main hall holds \(3^9\) books and a small reading room holds \(3^5\) books. The hall holds how many times as many books as the room? Give your answer as a power of \(3\).
Show the solution
How many times as many is a division. $$\dfrac{3^9}{3^5}=3^{9-5}=\boxed{3^4}$$ Dividing powers of the same base subtracts the exponents, so the huge \(3^9\) never has to be written out. The hall holds \(81\) times as many books.
Practice
Two identical servers each store \(2^{19}\) files. Put together, they store how many files in total? Write the total as a single power of \(2\). Careful, the answer is not \(4^{19}\).
Show the solution
Two equal stacks add to twice one stack. $$2^{19}+2^{19}=2\times2^{19}=\boxed{2^{20}}$$ Doubling lifts the exponent by one and leaves the base alone. It never turns into \(4\).
Practice
Watch which operation you are in. Multiplying two copies of \(2^{19}\) adds the exponents and lands far off at \(2^{38}\). Adding two copies only doubles. When equal powers are added you pull the number of copies out in front, so \(2^{19}+2^{19}=2\times2^{19}\), and only because that leading \(2\) matches the base does the exponent climb to \(2^{20}\).
Practice
A microscope magnifies by \(10^3\) at one setting. To study a single cell you stack that same setting \(4\) times in a row, each stage feeding its image into the next, for a magnification of \((10^3)^4\). Write that as a single power of \(10\).
Show the solution
Four stages each multiply by \(10^3\), so the total is \(10^3\) multiplied by itself four times, which is \((10^3)^4\). A power of a power multiplies the exponents. $$(10^3)^4=10^{3\times4}=\boxed{10^{12}}$$ A trillion-fold, written without a single zero.
Practice
A square mosaic is \(6\) tiles wide and \(6\) tiles tall. An artist refines it so that every tile is rebuilt as a \(5\)-by-\(5\) block of smaller tiles. How many small tiles fill the finished mosaic? Find it by writing the count as \((6\times5)^2\) and letting the exponent land on each factor.
Show the solution
Each side once spanned \(6\) tiles, and every tile is now \(5\) small tiles across, so a side runs \(6\times5=30\) small tiles. The mosaic holds \((6\times5)^2\) of them, and an exponent on a product drops onto each factor. $$(6\times5)^2=6^2\times5^2=36\times25=\boxed{900}$$ Squaring the product and multiplying the squares count the same tiles.
Practice
Three gears feed one counter. Gear A multiplies it by \(8\), gear B by \(4\), and gear C by \(2\). You crank gear A three times, gear B twice, and gear C once. By what single power of \(2\) has the counter grown?
Show the solution
Rewrite every gear in base \(2\). Cranking them gives $$8^3\times4^2\times2=(2^3)^3\times(2^2)^2\times2^1=2^9\times2^4\times2^1.$$ There is one base now, so add the exponents, \(9+4+1=14\), giving \(\boxed{2^{14}}\). All three gears were powers of \(2\), so a single base covers them all.
Practice
That is the move worth keeping. When the bases look different but are all powers of one number, rewrite every piece in that one base. The moment a single base rules the expression, the laws take over and the whole thing folds down to one tidy power.
Practice
Two students simplify the same thing and get opposite signs. One writes \((-2)^4\times2^3\), the other writes \(-2^4\times2^3\). Work out each value, then say how far apart the two results sit.
Show the solution
The parentheses decide everything. In \((-2)^4\) the sign is inside the power, and four negatives make a positive, so \((-2)^4=16\). In \(-2^4\) the exponent touches only the \(2\), and the minus waits outside, so \(-2^4=-(16)=-16\). Multiplying each by \(2^3=8\), $$16\times8=128,\qquad -16\times8=-128.$$ They sit \(128+128=\boxed{256}\) apart. The exponent grabs exactly what it touches, no more.
Practice
A puzzle claims that \(4^{12}\) and \(8^8\) are exactly the same size. Settle it by rewriting each as a power of \(2\). What single power of \(2\) do they share?
Show the solution
Send both to base \(2\). $$4^{12}=(2^2)^{12}=2^{24},\qquad 8^8=(2^3)^8=2^{24}$$ Same base and same exponent, so the puzzle is right and both equal \(\boxed{2^{24}}\). Once two numbers share a base, comparing them is just comparing exponents.
Practice
A number is built as \(2^6\times5^9\). How many digits does it have once it is written out in full?
Show the solution
A \(2\) and a \(5\) make a \(10\), but here the \(5\)s outnumber the \(2\)s. The six \(2\)s pair with six \(5\)s, leaving \(5^3\) on its own. $$2^6\times5^9=(2^6\times5^6)\times5^3=10^6\times125=125000000$$ That is \(125\) trailing six zeros, so \(\boxed{9}\) digits in all.
Practice
Every trailing zero on a whole number is a hidden \(10\), and every \(10\) is a \(2\) paired with a \(5\). That is the whole reason behind round numbers. To learn how many zeros end a giant product, you do not multiply it out, you just count how many \(2\)-and-\(5\) pairs are inside.
Practice
Collapse this tower into a single power of \(2\). $$\dfrac{8^6\times16^3}{4^5\times2^4}$$
Show the solution
Send every base to \(2\). On top, \(8^6=(2^3)^6=2^{18}\) and \(16^3=(2^4)^3=2^{12}\), so the numerator is \(2^{18}\times2^{12}=2^{30}\). On the bottom, \(4^5=(2^2)^5=2^{10}\) and \(2^4\) stays as it is, so the denominator is \(2^{10}\times2^4=2^{14}\). Dividing same bases subtracts the exponents. $$\dfrac{2^{30}}{2^{14}}=2^{30-14}=\boxed{2^{16}}$$ That is \(65536\). Once everything sits in one base, the laws flatten the whole tower.
Practice
This sum will not merge, because the powers are not equal. $$2^{10}+2^{11}+2^{12}$$ Find its exact value.
Show the solution
Factor out the smallest power, since all three terms contain \(2^{10}\). $$2^{10}+2^{11}+2^{12}=2^{10}(1+2+4)=2^{10}\times7=1024\times7=\boxed{7168}$$ Pulling out the lowest power is the move whenever a sum of powers will not collapse on its own.
Practice
Which is larger, \(2^{40}\) or \(3^{30}\)? Decide with no calculator by giving both a shared exponent.
Show the solution
Both exponents are multiples of \(10\), so give each number the exponent \(10\). $$2^{40}=(2^4)^{10}=16^{10},\qquad 3^{30}=(3^3)^{10}=27^{10}$$ The exponents match, so the larger base wins, and since \(27>16\) the bigger number is \(\boxed{3^{30}}\). When the bases will not match, a shared exponent does the same job.