QuanticaAlgebra IOpen in the courseMultiplication compresses repeated addition into one step. Exponents do the same for repeated multiplication. Every rule in this lesson comes from one habit, counting factors, and once the rules are in hand, requiring them to keep working is what gives zero, negative, and fractional exponents their values.
Problem
Ten 2s multiplied together is written \(2^{10}\). Multiplying them out one at a time works, but it is slow. Instead split the ten factors into two equal groups of five, compute one group, then finish with a single multiplication. What is \(2^{10}\)?
Show a hint
- One group of five 2s multiplies out to a number you know well.
Show the full solution
Five 2s multiply to \(32\), and the two groups give \(32\cdot 32=\boxed{1024}\). That is five multiplications instead of the nine you would do taking one factor at a time.
Problem
The product \(8\cdot 32\) is itself a power of 2. Each factor is itself a power of 2, so the whole product must be one too. Write \(8\cdot 32\) as \(2^x\) and enter \(x\).
Show a hint
- Count the factors of 2 in each piece before multiplying anything.
Show the full solution
\(8\) is three factors of 2 and \(32\) is five more, so the product is eight factors of 2 in one row and \(x=\boxed{8}\). Counting factors beats multiplying out to \(256\) and factoring back down.
Problem
Evaluate \(\frac{5^8}{5^5}\) without ever computing \(5^8\) or \(5^5\). Write the top and the bottom as strings of factors, cancel the shared ones, then multiply what is left.
Show a hint
- How many factors of 5 are left after the cancelling?
Show the full solution
The five factors of 5 on the bottom cancel five of the eight on top, leaving \(5\cdot 5\cdot 5=\boxed{125}\). Brute force gives \(390625\div 3125\) and the same answer, but it throws away the structure.
Problem
The expression \((7^2)^3\) means three factors of \(7^2\). Written out, it is a single power of 7, so \((7^2)^3=7^x\) for some \(x\). Count the total number of 7s and enter \(x\).
Show a hint
- Each copy of \(7^2\) contributes two 7s to the row.
Show the full solution
Three copies of \(7^2\), each holding two 7s, make \(2\cdot 3=6\) factors of 7, so \(x=\boxed{6}\). Adding to get \(7^5\) is the common slip, since the 3 counts copies rather than extra factors.
Problem
Walk down the powers of 3. \(3^4=81\), \(3^3=27\), \(3^2=9\), and each step down divides by 3. Keep the pattern going past \(3^1\) to \(3^0\). What value does the pattern force?
Show a hint
- What is 3 divided by 3?
Show the full solution
Each step down divides by 3, so \(3^2=9\) drops to \(3^1=3\), and one more step gives \(3\div 3\). So \(3^0=\boxed{1}\). The reflex answer of \(0\) never fits, since dividing 3 by 3 cannot land there.
Problem
The pattern does not stop at 1. Take two more steps down from \(3^0\), dividing by 3 at each step, to land on \(3^{-2}\). Enter its value as a fraction.
Show a hint
- Each step still divides by 3, even after the values drop below 1.
Show the full solution
One step below \(3^0=1\) is \(\frac{1}{3}\), and one more is \(\frac{1}{9}\), so \(3^{-2}=\boxed{\frac{1}{9}}\). A negative exponent makes a reciprocal, not a negative number, so \(-9\) is the wrong turn.
Problem
The product rule keeps working when an exponent is negative. Evaluate \(2^{-3}\cdot 2^7\) by adding the exponents before computing anything.
Show a hint
- Add the exponents first. Only then evaluate the single power that remains.
Show the full solution
Add the exponents, \(-3+7=4\), so the product is \(2^4=\boxed{16}\). The long way finds \(2^{-3}=\frac{1}{8}\) and \(2^7=128\) separately and then divides, landing in the same place with more work.
Problem
Suppose the power-of-a-power rule keeps working when an exponent is \(\frac{1}{2}\). Then \(\left(49^{1/2}\right)^2=49^{\frac{1}{2}\cdot 2}=49^1=49\), so \(49^{1/2}\) must be a number whose square is \(49\). What positive number is it?
Show a hint
- Which positive number, times itself, gives \(49\)?
Show the full solution
\(7\cdot 7=49\), so \(49^{1/2}=\boxed{7}\). Both \(7\) and \(-7\) square to \(49\), and the next block explains why the notation names the positive one.
Problem
Evaluate \(9^{3/2}\). The exponent offers two orders, root first or cube first. Both are legal, but only one keeps every number small.
Show a hint
- Write \(\frac{3}{2}\) as \(\frac{1}{2}\) applied first, then \(3\).
- What is \(9^{1/2}\)?
Show the full solution
Take the root first. \(9^{1/2}=3\), and \(3^3=\boxed{27}\). Cubing first is legal and gives the same answer, but it routes you through \(9^3=729\) before the square root, while root first keeps every number small.
One definition and one habit, counting factors, produced every rule here, and insisting those rules keep working gave \(a^0\), \(a^{-n}\), and \(a^{1/n}\) their meanings. Next, 1.6 returns to expressions and evaluates them at given values, and Chapter 3 brings variables into the laws you just built.
Practice these ideas
Practice
Warm up with the definition. \(5^4\) means four factors of \(5\) multiplied together, nothing more. Evaluate it as a single number.
Show the solution
Group the four \(5\)s into two pairs, each worth \(25\), so \(5^4 = 25\cdot 25 = \boxed{625}\). Multiplying one factor at a time gets there too, just with more chances to slip.
Practice
The product \(121 \cdot 11\) is itself a power of \(11\), since each of its two factors is one. Write the product as \(11^x\) and enter \(x\).
Show the solution
\(121 = 11^2\) and \(11 = 11^1\), so the product holds \(2 + 1 = 3\) factors of \(11\), and \(x = \boxed{3}\). The slip is entering \(1331\), the product itself, when the question asks for the exponent.
Practice
Evaluate \(\frac{12^9}{12^7}\). Cancel the shared factors before you multiply anything, and only then compute what is left.
Show the solution
The seven factors of \(12\) on the bottom cancel seven of the nine on top, leaving \(12^{9-7} = 12^2 = \boxed{144}\). Writing out \(12^9\), a ten-digit number, reaches the same answer the slow way.
Practice
\((17^4)^3\) collapses to a single power of \(17\). Count the total factors of \(17\) it contains and enter the exponent \(x\) in \((17^4)^3 = 17^x\).
Show the solution
Each of the three copies of \(17^4\) contributes four factors of \(17\), so there are \(4 \cdot 3 = 12\) in all and \(x = \boxed{12}\). Adding to get \(7\) confuses this with the product rule, where the factors sit side by side instead of in copies.
Practice
Evaluate \(14^0 \cdot 14^2\). One of the two factors is much simpler than it looks, so pin down its value before you multiply.
Show the solution
\(14^0 = 1\), so the product is \(1 \cdot 14^2 = \boxed{196}\). Any nonzero base to the zero power is \(1\), and the reflex answer of \(0\) is the usual miss.
Practice
Write \(5^{-2}\) as a fraction in lowest terms. Decide what the negative exponent actually does to the base before touching any arithmetic.
Show the solution
\(5^{-2} = \frac{1}{5^2} = \boxed{\frac{1}{25}}\). The minus in the exponent takes a reciprocal and never attaches a sign, so \(-25\) and \(-\frac{1}{25}\) are both wrong turns.
Practice
Evaluate \(15^{-4} \cdot 15^6\). Add the exponents before computing anything, then evaluate the small power that remains.
Show the solution
Add the exponents, \(-4 + 6 = 2\), so the product is \(15^2 = \boxed{225}\). Working out \(15^6 = 11390625\) and dividing by \(15^4\) gets there too, with far more arithmetic.
Practice
Evaluate \((-1)^{15} + (-1)^{16}\). Count the negative factors in each power, decide the sign each count forces, then add the two values.
Show the solution
Fifteen negative factors pair off with one left over, so \((-1)^{15} = -1\). Sixteen pair off exactly, so \((-1)^{16} = 1\). The sum is \(-1 + 1 = \boxed{0}\). An even exponent erases the sign, so a negative base does not always give a negative result.
Practice
Evaluate \(169^{1/2}\). The exponent names the one nonnegative number whose square is \(169\), so search for that number directly.
Show the solution
\(13 \cdot 13 = 169\), so \(169^{1/2} = \boxed{13}\). \(-13\) squares to \(169\) as well, but the notation names exactly one number and the convention keeps the nonnegative one.
Practice
Evaluate \((-27)^{1/3}\), the number whose cube is \(-27\). Decide first whether such a number can even exist, then find it.
Show the solution
\((-3)^3 = -27\), so \((-27)^{1/3} = \boxed{-3}\). An odd root of a negative exists because an odd count of negative factors stays negative, while an even root of a negative names no real number.
Practice
Evaluate \(256^{3/4}\). Take the fourth root first and cube the result, keeping every number in the computation small along the way.
Show the solution
\(256^{1/4} = 4\) since \(4^4 = 256\), and then \(4^3 = \boxed{64}\). Cubing first is legal but sends you through \(256^3 = 16777216\) before the fourth root.
Practice
Find the number \(x\) that makes \(4^x = 8\) true. The powers of \(4\) you know jump right past \(8\), yet an \(x\) exists. Enter it as a fraction.
Show the solution
Write both sides in base \(2\). \(4^x = 2^{2x}\) and \(8 = 2^3\), so \(2x = 3\) and \(x = \boxed{\frac{3}{2}}\). No whole number works here, since \(4^1 = 4\) and \(4^2 = 16\) straddle \(8\).