Roots and powers

Recall that the fourth power of 55 is defined as

54=5⋅5⋅5⋅5⏟4 times=6255^4 = \underbrace{5\cdot 5\cdot 5\cdot 5}_{4 \text{ times}} = 625

and if the power is negative, we get fractions:

5−4=154=16255^{-4} = \frac{1}{5^4}=\frac{1}{625}

This makes sense, if we want to keep up a certain regularity:

54=625↓:553=125↓:552=25↓:551=5↓:550=1↓:55−1=15=151↓:55−2=125=152↓:55−3=1125=153...\begin{array}{rll} 5^4 & = & 625\\ & \downarrow :5 & \\ 5^3 & = & 125\\ & \downarrow :5 & \\ 5^2 & = & 25\\ & \downarrow :5 & \\ 5^1 & = & 5\\ & \downarrow :5 & \\ 5^0 & = & 1\\ & \downarrow :5 & \\ 5^{-1} & = & \frac{1}{5}=\frac{1}{5^1}\\ & \downarrow :5 & \\ 5^{-2} & = & \frac{1}{25}=\frac{1}{5^2}\\ & \downarrow :5 & \\ 5^{-3} & = & \frac{1}{125}=\frac{1}{5^3}\\ & ... & \end{array}

Also note that it follows 50=15^0=1. More generally, we have the following power rules:

Theorem 1: Power rules (or power laws)

For any numbers a,b,na, b, n and mm it is:

D:an=a⋅...⋅a⏟n times0:a0=1R:a1/n=an and ap/q=apq=(aq)pF:a−n=1anSB:an⋅am=an+m and anam=an−mSE:an⋅bn=(ab)n and anbn=(ab)nPP:(an)m=anm\begin{array}{rrllll} \text{D:} & a^n &=&\underbrace{a\cdot ...\cdot a}_{n \text{ times}} & &\\[0.4em] \text{0:} & a^0&=&1 &\\[0.3em] \text{R:} & a^{1/n} &=& \sqrt[n]{a} & \text{ and } & a^{p/q}= \sqrt[q]{a^p} = \left(\sqrt[q]{a}\right)^p\\[0.3em] \text{F:} & a^{-n} &=& \frac{1}{a^n} & & \\ & & & & & \\ \text{SB:} & a^n\cdot a^m &=& a^{n+m} & \text{ and } & \frac{a^n}{a^m}=a^{n-m}\\[0.3em] \text{SE:} & a^n\cdot b^n &=& (ab)^n & \text{ and } & \frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n\\[0.3em] \text{PP:} & (a^n)^m &=& a^{nm} & & \end{array}

where D=definition, 0=zero power law, R=root law, F=fraction law, SB=same base law, SE=same exponent law, and PP=power of a power law. One possible way to remember: "D0RFBEP".

Observe the definition for rational exponents, which is new to you:

a1/n=ana^{1/n}=\sqrt[n]{a}

So according to this rule, we have

251/2=252=525^{1/2}=\sqrt[2]{25}=5

and this is indeed so because

(251/2)2=251=25(25^{1/2})^2=25^1=25

The same argument can be made in general. We have a1/n=ana^{1/n}=\sqrt[n]{a} because

(a1/n)n=a1=a(a^{1/n})^n = a^1 = a

To find the value of, say 251225^\frac{1}{2} in your head, we can first convert the power into a root, and then calculate the root: 2512=25=525^\frac{1}{2}=\sqrt{25}=5. But often we simply use the calculator by directly using the key x□\boxed{x^\square}.

Example 1

Determine without calculator:

  1. 81/3=28^{1/3}=2, because 81/3=838^{1/3}=\sqrt[3]{8}
  2. 8−1/3=128^{-1/3}=\frac{1}{2}, because 8−1/3=181/3=183=128^{-1/3}=\frac{1}{8^{1/3}}=\frac{1}{\sqrt[3]{8}}=\frac{1}{2}
  3. 82/3=48^{2/3}=4, because 82/3=81/3⋅2=(81/3)2=(83)2=22=48^{2/3}=8^{1/3 \cdot 2}=\left(8^{1/3}\right)^2=\left(\sqrt[3]{8}\right)^2=2^2=4

Verify the results using the calculator.

Exercise 1
  1. Use the power rules to show that
    1. 7⋅83=73⋅83\sqrt[3]{7\cdot 8} = \sqrt[3]{7}\cdot \sqrt[3]{8}
    2. 783=7383\sqrt[3]{\frac{7}{8}} = \frac{\sqrt[3]{7}}{\sqrt[3]{8}}
    3. 7103=710/3\sqrt[3]{7^{10}}=7^{10/3}
    4. 733=7\sqrt[3]{7^3}=7
  2. Use the power rule to show that the following rules are correct:
    1. a⋅bn=an⋅bn\sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot \sqrt[n]{b}
    2. abn=anbn\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}
    3. amn=am/n\sqrt[n]{a^m}=a^{m/n}
    4. ann=a\sqrt[n]{a^n}=a
    5. an⋅am=an+mn⋅m\sqrt[n]{a}\cdot \sqrt[m]{a}=\sqrt[n\cdot m]{a^{n+m}}
  3. Use the calculator to show that the following is not correct: 7+83=73+83\sqrt[3]{7+ 8} = \sqrt[3]{7}+\sqrt[3]{8}
  4. Use two different methods on the calculator to calculate

    1. 410\sqrt[10]{4}

    2. 0.755\sqrt[5]{0.75}

Solution
  1. It is
    1. 7⋅83=D(7⋅8)1/3=SE71/3⋅81/3=D73⋅83\sqrt[3]{7\cdot 8} \overset{D}{=} (7\cdot 8)^{1/3} \overset{SE}{=} 7^{1/3}\cdot 8^{1/3}\overset{D}{=}\sqrt[3]{7}\cdot \sqrt[3]{8}
    2. 783=D(78)1/3=SE(71/381/3)=D7383\sqrt[3]{\frac{7}{8}} \overset{D}{=} \left(\frac{7}{8}\right)^{1/3} \overset{SE}{=} \left(\frac{7^{1/3}}{8^{1/3}}\right) \overset{D}{=} \frac{\sqrt[3]{7}}{\sqrt[3]{8}}
    3. 7103=D(710)1/3=PP710/3\sqrt[3]{7^{10}}\overset{D}{=} \left(7^{10}\right)^{1/3}\overset{PP}{=} 7^{10/3}
    4. Follows from c: 733=73/3=7\sqrt[3]{7^3}=7^{3/3}=7
  2. It is
    1. a⋅bn=D(a⋅b)1/n=SEa1/n⋅b1/n=Dan⋅bn\sqrt[n]{a\cdot b} \overset{D}{=} (a\cdot b)^{1/n} \overset{SE}{=} a^{1/n}\cdot b^{1/n}\overset{D}{=}\sqrt[n]{a}\cdot \sqrt[n]{b}
    2. abn=D(ab)1/n=SE(a1/nb1/n)=Danbn\sqrt[n]{\frac{a}{b}} \overset{D}{=} \left(\frac{a}{b}\right)^{1/n} \overset{SE}{=} \left(\frac{a^{1/n}}{b^{1/n}}\right) \overset{D}{=} \frac{\sqrt[n]{a}}{\sqrt[n]{b}}
    3. amn=D(am)1/n=PPam/n\sqrt[n]{a^{m}}\overset{D}{=} \left(a^{m}\right)^{1/n}\overset{PP}{=} a^{m/n}
    4. Follows from c: ann=an/n=a\sqrt[n]{a^n}=a^{n/n}=a
    5. an⋅am=Da1/na1/m=SBa1/n+1/m=an+mn⋅m=Dan+mn⋅m\sqrt[n]{a}\cdot \sqrt[m]{a}\overset{D}{=}a^{1/n} a^{1/m} \overset{SB}{=} a^{1/n+1/m}=a^\frac{n+m}{n\cdot m}\overset{D}{=}\sqrt[n\cdot m]{a^{n+m}}
  3. not shown ...
  4. Method 1 is with the root key: 2nd and x□\boxed{2nd} \text{ and } \boxed{x^\square}, method 2 is to convert the root into a power, and use key x□\boxed{x^\square}.
Exercise 2

  1. Determine without calculator, and write the result as a natural number or a fraction:

    1. 811/281^{1/2}
    2. 10 0001/410\,000^{1/4}
    3. 11/31^{1/3}
    4. 01/50^{1/5}
    5. 271/327^{1/3}
    6. 6250.5625^{0.5}
    7. 2560.25256^{0.25}
    8. (27125)1/3\left(\frac{27}{125}\right)^{1/3}
    9. (110 000)1/4\left(\frac{1}{10\,000}\right)^{1/4}
    10. 25−1/225^{-1/2}
    11. 343−1/3343^{-1/3}
    12. 16−1/416^{-1/4}
    13. 243−1/5243^{-1/5}
    14. 49−0.549^{-0.5}
    15. 256−0.125256^{-0.125}
    16. 82/38^{2/3}
    17. 0.25−0.50.25^{-0.5}
    18. 160.7516^{0.75}
    19. (165)0.1(16^5)^{0.1}
    20. 5⋅320.45 \cdot 32^{0.4}

  2. Convert to a power of the form ana^n where the base aa is a natural number (as small as possible):

    1. 534\sqrt[4]{5^3}
    2. 10 0006\sqrt[6]{10\,000}
    3. 0.00017\sqrt[7]{0.0001}
    4. 0.53\sqrt[3]{0.5}
    5. 0.125\sqrt{0.125}
    6. 1814\frac{1}{\sqrt[4]{81}}

  3. Write as a single root an\sqrt[n]{a} where the radicand aa is a natural number:

    1. 11123⋅11135⋅11130\sqrt[3]{111^2} \cdot \sqrt[5]{111^3} \cdot \sqrt[30]{111}
    2. 514623512346\frac{\sqrt[23]{51^{46}}}{\sqrt[46]{51^{23}}}

  4. Determine without calculator:

    1. 161/2+82/3+363/2−1251/3+272/316^{1/2}+8^{2/3}+36^{3/2}-125^{1/3}+27^{2/3}
    2. 10002/3−1252/3+161/2+82/3−813/4+272/31000^{2/3}-125^{2/3}+16^{1/2}+8^{2/3}-81^{3/4}+27^{2/3}

  5. Sort by decreasing order (without calculator):

    1. 49,490,49−1,491,491.5,49−1.549, 49^0, 49^{-1}, 49^1, 49^{1.5}, 49^{-1.5}
    2. 625−0.75,144−1.5,2−10,100−1.5625^{-0.75}, 144^{-1.5}, 2^{-10}, 100^{-1.5}

  6. Determine without calculator

    1. 7296−31255+12964−3433\sqrt[6]{729}-\sqrt[5]{3125}+\sqrt[4]{1296}-\sqrt[3]{343}

    2. 1325−11253+1646−16254\sqrt[5]{\frac{1}{32}}-\sqrt[3]{\frac{1}{125}}+\sqrt[6]{\frac{1}{64}}-\sqrt[4]{\frac{1}{625}}

    3. 64−16−243−15+625−14\sqrt[6]{64^{-1}}-\sqrt[5]{243^{-1}}+\sqrt[4]{625^{-1}}

Solution

  1. It is

    1. 811/2=(92)1/2=91=981^{1/2}=(9^2)^{1/2}=9^1=9
    2. 10 0001/4=(104)1/4=101=1010\,000^{1/4}=(10^4)^{1/4}=10^1=10
    3. 11/3=13=11^{1/3}=\sqrt[3]{1}=1
    4. 01/5=05=00^{1/5}=\sqrt[5]{0}=0
    5. 271/3=(33)1/3=31=327^{1/3}=(3^3)^{1/3}=3^1=3
    6. 6250.5=(252)0.5=251=25625^{0.5}=(25^2)^{0.5}=25^1=25
    7. 2560.25=(28)0.25=22=4256^{0.25}=(2^8)^{0.25}=2^2=4
    8. (27125)1/3=271/31251/3=35\left(\frac{27}{125}\right)^{1/3}=\frac{27^{1/3}}{125^{1/3}}=\frac{3}{5}
    9. (110 000)1/4=11/4(104)1/4=110\left(\frac{1}{10\,000}\right)^{1/4}=\frac{1^{1/4}}{(10^4)^{1/4}}=\frac{1}{10}
    10. 25−1/2=(52)−1/2=5−1=1525^{-1/2}=(5^2)^{-1/2}=5^{-1}=\frac{1}{5}
    11. 343−1/3=(73)−1/3=7−1=17343^{-1/3}=(7^3)^{-1/3}=7^{-1}=\frac{1}{7}
    12. 16−1/4=(24)−1/4=2−1=1216^{-1/4}=(2^4)^{-1/4}=2^{-1}=\frac{1}{2}
    13. 243−1/5=(35)−1/5=3−1=13243^{-1/5}=(3^5)^{-1/5}=3^{-1}=\frac{1}{3}
    14. 49−0.5=(72)−0.5=7−1=1749^{-0.5}=(7^2)^{-0.5}=7^{-1}=\frac{1}{7}
    15. 256−0.125=(28)−0.125=2−1=12256^{-0.125}=(2^8)^{-0.125}=2^{-1}=\frac{1}{2}
    16. 82/3=(23)2/3=22=48^{2/3}=\left(2^3\right)^{2/3}=2^2=4
    17. 0.25−0.5=10.25=10.5=20.25^{-0.5}=\frac{1}{\sqrt{0.25}}=\frac{1}{0.5}=2
    18. 160.75=163/4=(24)3/4=23=816^{0.75}=16^{3/4}=\left(2^4\right)^{3/4}=2^3=8
    19. (165)0.1=161/2=16=4(16^5)^{0.1}=16^{1/2}=\sqrt{16}=4
    20. 5⋅320.4=5⋅322/5=5⋅(25)2/5=5⋅22=205 \cdot 32^{0.4}=5\cdot 32^{2/5}=5\cdot \left(2^5\right)^{2/5}=5\cdot 2^2=20

  2. It is

    1. 534=53/4\sqrt[4]{5^3}=5^{3/4}
    2. 10 0006=1046=102/3\sqrt[6]{10\,000}=\sqrt[6]{10^4}=10^{2/3}
    3. 0.00017=10−47=10−4/7\sqrt[7]{0.0001}=\sqrt[7]{10^{-4}}=10^{-4/7}
    4. 0.53=2−13=2−1/3\sqrt[3]{0.5}=\sqrt[3]{2^{-1}}=2^{-1/3}
    5. 0.125=18=2−32=2−3/2\sqrt{0.125}=\sqrt{\frac{1}{8}}=\sqrt[2]{2^{-3}}=2^{-3/2}
    6. 1814=1344=13=3−1\frac{1}{\sqrt[4]{81}}=\frac{1}{\sqrt[4]{3^4}}=\frac{1}{3}=3^{-1}

  3. It is

    1. 11123⋅11135⋅11130=1112/3⋅1113/5⋅1111/30=1112/3+3/5+1/30=11113/10\begin{array}{l}\sqrt[3]{111^2} \cdot \sqrt[5]{111^3} \cdot \sqrt[30]{111}\\ = 111^{2/3}\cdot 111^{3/5}\cdot 111^{1/30}\\ = 111^{2/3+3/5+1/30}\\ = 111^{13/10} \end{array}
    2. 514623512346=5146/235123/46=512511/2=513/2\frac{\sqrt[23]{51^{46}}}{\sqrt[46]{51^{23}}}=\frac{51^{46/23}}{51^{23/46}}=\frac{51^2}{51^{1/2}}=51^{3/2}

  4. It is

    1. 161/2+82/3+363/2−1251/3+272/3=(42)1/2+(23)2/3+(62)3/2−(53)1/3+(33)2/3=4+22+63−5+32=4+4+216−5+9=228\begin{array}{l} 16^{1/2}+8^{2/3}+36^{3/2}-125^{1/3}+27^{2/3}\\= (4^2)^{1/2}+(2^3)^{2/3}+(6^2)^{3/2}-(5^3)^{1/3}+(3^3)^{2/3}\\=4+2^2+6^3-5+3^2\\= 4+4+216-5+9=228\end{array}
    2. 10002/3−1252/3+161/2+82/3−813/4+272/3=(103)2/3−(53)2/3+(42)1/2+(23)2/3−(34)3/4+(33)2/3=102−52+4+22−33+32=100−25+4+4−27+9=65\begin{array}{l} 1000^{2/3}-125^{2/3}+16^{1/2}+8^{2/3}-81^{3/4}+27^{2/3}\\ = (10^3)^{2/3}-(5^3)^{2/3}+(4^2)^{1/2}+(2^3)^{2/3}-(3^4)^{3/4}+(3^3)^{2/3}\\= 10^2-5^2+4+2^2-3^3+3^2\\= 100-25+4+4-27+9\\=65\end{array}

  5. It is

    1. 491.5>49=491>490>49−1>49−1.549^{1.5}> 49=49^1> 49^0> 49^{-1}> 49^{-1.5}

    2. 625−0.75=1/53=1/125144−1.5=1/123=1/17282−10=1/210=1/1024100−1.5=1/103=1/1000\begin{array}{lllll} 625^{-0.75}&=&1/5^3&=&1/125\\ 144^{-1.5}&=&1/12^3&=&1/1728\\ 2^{-10}&=&1/2^{10}&=&1/1024\\ 100^{-1.5}&=&1/10^3&=&1/1000 \end{array}

      Thus 625−0.75>100−1.5>2−10>144−1.5625^{-0.75}> 100^{-1.5}> 2^{-10}> 144^{-1.5}

  6. We have

    1. 7296−31255+12964−3433=3−5+6−7\sqrt[6]{729}-\sqrt[5]{3125}+\sqrt[4]{1296}-\sqrt[3]{343}=3-5+6-7
    2. 1325−11253+1646−16254=12−15+12−15=35\sqrt[5]{\frac{1}{32}}-\sqrt[3]{\frac{1}{125}}+\sqrt[6]{\frac{1}{64}}-\sqrt[4]{\frac{1}{625}}=\frac{1}{2}-\frac{1}{5}+\frac{1}{2}-\frac{1}{5}=\frac{3}{5}
    3. 64−16−243−15+625−14=12−13+15=1130\sqrt[6]{64^{-1}}-\sqrt[5]{243^{-1}}+\sqrt[4]{625^{-1}}=\frac{1}{2}-\frac{1}{3}+\frac{1}{5}=\frac{11}{30}