QuanticaAlgebra IOpen in the courseNothing in the 3.1 laws ever asked an exponent to be a whole number, so \(x^{1/2}\) is legal to write, and so far it means nothing. The plan is the one that gave \(x^{0}\) and \(x^{-3}\) their meanings in 3.2. Refuse to guess, run the laws on the new symbol, and keep whichever value they force.
Problem
Whatever \(21^{1/2}\) turns out to mean, power of a power from 3.1 still applies to it. Apply that law to \(\left(21^{1/2}\right)^{2}\) and enter the value.
Show a hint
- You do not need to know what \(21^{1/2}\) is. Power of a power multiplies the exponents.
- \(\frac{1}{2}\cdot 2=1\), and a first power is just the base.
Show the full solution
Power of a power multiplies the exponents, so $$\left(21^{1/2}\right)^{2}=21^{\frac{1}{2}\cdot 2}=21^{1}=\boxed{21}.$$ Whatever number \(21^{1/2}\) is, its square has to be \(21\), and \(21\) is not even a perfect square. The law never needed to know the value.
Problem
Here is a second route, independent of the first. The product rule from 3.1 adds exponents, so \(169^{1/2}\cdot 169^{1/2}=169^{\frac{1}{2}+\frac{1}{2}}=169^{1}\). So \(169^{1/2}\) times itself is \(169\). Enter the value of \(169^{1/2}\).
Show a hint
- The exponents add to \(1\), so the product is \(169\) itself. You are looking for a number that multiplies by itself to make \(169\).
- Its square ends in \(9\), so the number has to end in \(3\) or \(7\).
Show the full solution
The exponents add to \(1\), so \(169^{1/2}\cdot 169^{1/2}=169\), and \(13\cdot 13=169\), so $$169^{1/2}=\boxed{13}.$$ A different exponent law from the last problem gives the same conclusion. \(169^{1/2}\) has to be a number whose square is \(169\).
Problem
The unfinished business first. Two different numbers have a square equal to \(121\). Enter the negative one of the two.
Show a hint
- Find a number whose square is \(121\) first. There is an obvious positive one.
- \(11^{2}=121\), and one other number's square is also \(121\). Squaring wipes out a minus sign.
Show the full solution
\((-11)^{2}=(-11)(-11)=121\), so the negative candidate is \(\boxed{-11}\). Both \(11\) and \(-11\) square to \(121\), and the next block settles which of the two the symbol \(121^{1/2}\) names.
Problem
The last two blocks each ruled out a tempting wrong answer, and this problem offers both of them. Enter the value of \(400^{1/2}\).
Show a hint
- Not half of \(400\). The exponent asks what number squares to \(400\).
- Two numbers square to \(400\). By convention \(400^{1/2}\) means the nonnegative one.
Show the full solution
\(20^{2}=400\), so \(400^{1/2}=\boxed{20}\). The answer \(200\) treats the exponent as a factor, and \(-20\) also squares to \(400\), but by convention \(400^{1/2}\) means the nonnegative choice.
Problem
Nothing in the two routes needed the \(2\). Run the product rule with three copies, \(216^{1/3}\cdot 216^{1/3}\cdot 216^{1/3}=216^{1}\), so \(216^{1/3}\) is the number whose cube is \(216\). Enter it.
Show a hint
- The three exponents add to \(1\), so this is three equal factors multiplying to \(216\).
- Try small whole numbers. \(5^{3}=125\) is too small.
Show the full solution
\(6^{3}=6\cdot 6\cdot 6=216\), so \(216^{1/3}=\boxed{6}\). This time only one number works. \((-6)^{3}=-216\), not \(216\), so there was no tie between candidates to break.
Problem
The base is negative in both of \((-343)^{1/3}\) and \((-343)^{1/2}\), and exactly one of the two names a real number. Enter the value of the one that does.
Show a hint
- Ask what sign an even power of a real number can have, and what sign an odd power can have.
- No real number squares to a negative. Cubes are different, so look for a number whose cube is \(-343\).
Show the full solution
\((-7)^{3}=(-7)(-7)(-7)=-343\), so \((-343)^{1/3}=\boxed{-7}\). No real number squares to \(-343\), since squares are never negative, so \((-343)^{1/2}\) has no value. An odd root of a negative is negative, not missing.
Problem
Power of a power reads \(64^{2/3}\) two ways, since \(\frac{1}{3}\cdot 2\) and \(2\cdot\frac{1}{3}\) are the same exponent. So \(64^{2/3}\) is both \(\left(64^{1/3}\right)^{2}\) and \(\left(64^{2}\right)^{1/3}\). Work out both. They agree, and that value is the answer.
Show a hint
- One reading takes the cube root first and then squares. The other squares first and then takes the cube root.
- The root-first route is \(4^{2}\). The power-first route has to walk through \(64^{2}=4096\).
Show the full solution
Root first, \(64^{1/3}=4\) and \(4^{2}=16\). Power first, \(64^{2}=4096\), and \(16^{3}=4096\), so \(4096^{1/3}=16\) as well. Both readings give $$64^{2/3}=\boxed{16}.$$ One order stays small the whole way. The other detours through \(4096\) for the identical answer.
Problem
Evaluate \(256^{3/8}\). The eighth root is invisible until you write the base as a power of something small. Enter the value.
Show a hint
- Try writing \(256\) as a power of a small number first.
- \(256=2^{8}\), and then power of a power collapses the whole thing.
Show the full solution
\(256=2^{8}\), so $$256^{3/8}=\left(2^{8}\right)^{3/8}=2^{8\cdot\frac{3}{8}}=2^{3}=\boxed{8}.$$ Rewriting the base turned an invisible eighth root into one line of arithmetic.
Problem
A letter base changes none of this. For positive \(w\), the product \(w^{5/6}\cdot w^{13/6}\) collapses to a single power \(w^{k}\). Enter \(k\).
Show a hint
- The product rule from 3.1 still adds the exponents, fractions included.
- \(\frac{5}{6}+\frac{13}{6}\) has a whole-number value.
Show the full solution
The product rule adds the exponents. $$\frac{5}{6}+\frac{13}{6}=\frac{18}{6}=3,$$ so \(w^{5/6}\cdot w^{13/6}=w^{3}\) and \(k=\boxed{3}\). Multiplying the exponents instead applies power of a power where the product rule belongs.
Problem
All three moves in one problem, the reciprocal, the root and the power. Evaluate \(196^{-3/2}\). Enter the value as a fraction in lowest terms.
Show a hint
- Handle the minus the 3.2 way first. It puts \(196^{3/2}\) under a \(1\).
- \(196^{1/2}=14\), and then the \(3\) cubes it.
Show the full solution
The minus is a reciprocal, so \(196^{-3/2}=\frac{1}{196^{3/2}}\). Root first, \(196^{1/2}=14\), then \(14^{3}=2744\), so $$196^{-3/2}=\boxed{\frac{1}{2744}}.$$ Reading the minus as a sign on the value gives \(-2744\), wrong in both size and meaning.
Problem
Evaluate \(\left(\frac{32}{243}\right)^{2/5}\). A power of a fraction raises the top and the bottom separately, the way each base kept its own exponent in 3.1. Enter the value as a fraction in lowest terms.
Show a hint
- Take the fifth root of the top and of the bottom first.
- \(2^{5}=32\) and \(3^{5}=243\), and then each root gets squared.
Show the full solution
\(32^{1/5}=2\) since \(2^{5}=32\), and \(243^{1/5}=3\) since \(3^{5}=243\). Squaring each root gives $$\left(\frac{32}{243}\right)^{2/5}=\frac{2^{2}}{3^{2}}=\boxed{\frac{4}{9}}.$$ Root first kept every number in the work a single digit.
Problem
One equation to close. Find the exponent \(k\) with \(81^{k}=729\). Write both sides as powers of \(3\) first. Enter \(k\) as a fraction in lowest terms.
Show a hint
- \(81\) and \(729\) are both powers of \(3\).
- \(81^{k}=\left(3^{4}\right)^{k}=3^{4k}\), and the right side is \(3^{6}\).
Show the full solution
\(81=3^{4}\) and \(729=3^{6}\), so the equation is \(3^{4k}=3^{6}\), which forces \(4k=6\) and $$k=\boxed{\frac{3}{2}}.$$ Check it root first, \(81^{3/2}=\left(81^{1/2}\right)^{3}=9^{3}=729\). The lesson ends where it began, on an equation only a fractional exponent can answer.
Every root in this lesson was written as an exponent, and every answer came out an integer or a fraction, because every base was a perfect power. Roots come up often enough to have earned a symbol of their own. 3.4 introduces it, along with the question this lesson dodged, what happens when the base is not a perfect power.
Practice these ideas
Practice
You do not need to know what \(23^{1/2}\) is to answer this. Enter the value of \((23^{1/2})^{2}\).
Show the solution
Power of a power multiplies the exponents, so \((23^{1/2})^{2}=23^{(1/2)\cdot 2}=23^{1}=\boxed{23}\). The square of \(23^{1/2}\) is \(23\) by definition, and the law gives that without the value of \(23^{1/2}\) ever being needed.
Practice
Evaluate \(484^{1/2}\).
Show the solution
\(22^{2}=484\), so \(484^{1/2}=\boxed{22}\). The exponent asks for the nonnegative number whose square is \(484\), and \(22\) is the one that checks.
Practice
A student reads the \(\frac{1}{2}\) in \(324^{1/2}\) as a factor and answers \(162\). Enter the actual value.
Show the solution
\(18^{2}=324\), so \(324^{1/2}=\boxed{18}\). Multiplying by \(\frac{1}{2}\) gives \(162\), but the exponent asks what squares to \(324\), and \(162^{2}\) is nowhere near it.
Practice
A student answers \(-31\) for \(961^{1/2}\) and runs the check \((-31)^{2}=961\), which comes out right. The answer is still wrong. Enter the value of \(961^{1/2}\).
Show the solution
\(31^{2}=961\), so \(961^{1/2}=\boxed{31}\). Both \(31\) and \(-31\) square to \(961\), and the symbol names the nonnegative one by convention, so no amount of checking by squaring can rescue \(-31\). That candidate is written \(-(961^{1/2})\).
Practice
Evaluate \(6859^{1/3}\).
Show the solution
\(19^{3}=6859\), so \(6859^{1/3}=\boxed{19}\). The bounds \(10^{3}=1000\) and \(20^{3}=8000\) trap the root in the teens, and only a units digit of \(9\) cubes to something ending in \(9\), so \(19\) was the only candidate worth testing.
Practice
Evaluate \((-4913)^{1/3}\).
Show the solution
\(17^{3}=4913\), so \((-17)^{3}=-4913\) and \((-4913)^{1/3}=\boxed{-17}\). Three negative factors leave a negative product, so an odd root of a negative number is negative, not missing.
Practice
Of \((-361)^{1/2}\), \((-3375)^{1/3}\), \((-49)^{1/4}\), and \((-1)^{1/5}\), how many name a real number? Enter the count.
Show the solution
The even roots fail, since no real number raised to the power \(2\) or \(4\) is negative. The odd roots are real, \((-3375)^{1/3}=-15\) and \((-1)^{1/5}=-1\). That makes \(\boxed{2}\) real values. Parity settles all four before any arithmetic starts.
Practice
For positive \(z\), \(z^{3/4}\cdot z^{13/4}=z^{k}\). Enter \(k\).
Show the solution
The product rule adds the exponents, \(\frac{3}{4}+\frac{13}{4}=\frac{16}{4}=\boxed{4}\). Multiplying them instead is power of a power's move, and that law belongs to a different expression.
Practice
For positive \(c\), \((c^{5/12})^{18}=c^{k}\). Enter \(k\) as a fraction in lowest terms.
Show the solution
Power of a power multiplies the exponents, \(\frac{5}{12}\cdot 18=\frac{90}{12}=\boxed{\frac{15}{2}}\). Nothing requires the exponent to come out whole, so \(\frac{15}{2}\) is the finished answer.
Practice
For positive \(v\), \(\frac{v^{2/5}}{v^{22/5}}=v^{k}\). Enter \(k\).
Show the solution
The quotient rule subtracts, \(\frac{2}{5}-\frac{22}{5}=-\frac{20}{5}=\boxed{-4}\). Subtracting in the wrong order gives \(4\), off by exactly a sign, which is why the rule fixes the order as top minus bottom.
Practice
Evaluate \(900^{3/2}\). Take the root first to keep the numbers small.
Show the solution
\(900^{1/2}=30\) and \(30^{3}=27000\), so \(900^{3/2}=\boxed{27000}\). Power first lands in the same place, but it runs \(900^{3}\), a nine digit number, through the arithmetic on the way.
Practice
Evaluate \(\left(\frac{576}{841}\right)^{1/2}\). Enter a fraction in lowest terms.
Show the solution
The exponent applies to top and bottom separately, \(576^{1/2}=24\) and \(841^{1/2}=29\), so the value is \(\boxed{\frac{24}{29}}\). It is already in lowest terms, since \(29\) is prime and does not divide \(24\).
Practice
Enter \(676^{-1/2}\) as a fraction in lowest terms.
Show the solution
The minus flips first, \(676^{-1/2}=\frac{1}{676^{1/2}}\), and \(26^{2}=676\), so the value is \(\boxed{\frac{1}{26}}\). The minus makes a reciprocal, not a negative value, so the answer is a small positive fraction, never \(-26\).
Practice
Evaluate \(1024^{7/10}\).
Show the solution
\(1024=2^{10}\), so \(1024^{7/10}=(2^{10})^{7/10}=2^{7}=\boxed{128}\). Rewriting the base does all the work. The tenth root of \(1024\) was never something to hunt for directly.
Practice
Enter \(6561^{-3/8}\) as a fraction in lowest terms.
Show the solution
\(6561=3^{8}\), so \(6561^{3/8}=(3^{8})^{3/8}=3^{3}=27\), and the minus flips that to \(\boxed{\frac{1}{27}}\). The minus is 3.2's reciprocal move, the fraction is this lesson's root and power, and they stack without interfering.
Practice
Exactly one of \(-(784^{1/2})\) and \((-784)^{1/2}\) names a real number. Enter its value.
Show the solution
\(28^{2}=784\), so \(784^{1/2}=28\) and \(-(784^{1/2})=\boxed{-28}\). The other expression puts the minus inside the base, and no real number squares to \(-784\), so \((-784)^{1/2}\) has no value at all. The parentheses decide which computation you are being asked to run.
Practice
Find the exponent \(k\) with \(100^{k}=100000\). Write both sides as powers of \(10\) first. Enter \(k\) as a fraction in lowest terms.
Show the solution
\(100=10^{2}\) and \(100000=10^{5}\), so the equation reads \((10^{2})^{k}=10^{5}\), which is \(10^{2k}=10^{5}\). Matching exponents forces \(2k=5\), so \(k=\boxed{\frac{5}{2}}\). Check it root first, \(100^{5/2}=(100^{1/2})^{5}=10^{5}=100000\). No whole number works here, which is exactly the kind of equation fractional exponents exist to answer.