Express each of the following in exponential form
(a) \( \sqrt[7]{100} = 100^{1/7} \)
(b) \( \sqrt[4]{34} = 34^{1/4} \)
(c) \( \sqrt[11]{25^2} = (25^2)^{1/11} = 25^{2/11} \)
(d) \( \sqrt[7]{29^2} = (29^2)^{1/7} = 29^{2/7} \)
(e) \( \sqrt[7]{\dfrac{10}{7}} = \left(\dfrac{10}{7}\right)^{1/7} \)
(f) \( \sqrt[12]{111^3} = (111^3)^{1/12} = 111^{3/12} = 111^{1/4} \)
(g) \( \sqrt[3]{2^{-6}} = (2^{-6})^{1/3} = 2^{-2} \)
(h) \( \sqrt[4]{\left(\dfrac{7}{8}\right)^{-5}} = \left(\dfrac{7}{8}\right)^{-5/4} \)
Express each of the following as radicals
(a) \( 7^{1/5} = \sqrt[5]{7} \)
(b) \( 21^{1/8} = \sqrt[8]{21} \)
(c) \( (29)^{2/3} = \sqrt[3]{29^2} = \sqrt[3]{841} \)
(d) \( (335)^{7/5} = \sqrt[5]{335^7} \)
(e) \( \left(\dfrac{8}{15}\right)^{1/9} = \sqrt[9]{\dfrac{8}{15}} \)
(f) \( \left(\dfrac{17}{327}\right)^{3/7} = \sqrt[7]{\left(\dfrac{17}{327}\right)^3} \)
(g) \( \left(\dfrac{17}{21}\right)^{2/7} = \sqrt[7]{\left(\dfrac{17}{21}\right)^2} \)
(h) \( \left(\dfrac{1}{16}\right)^{1/5} = \sqrt[5]{\dfrac{1}{16}} \)
Express each of the following with positive indices
(a) \( x^{-4} = \dfrac{1}{x^4} \)
(b) \( x^{-1/3} = \dfrac{1}{x^{1/3}} \)
(c) \( x^{-2/5} = \dfrac{1}{x^{2/5}} \)
(d) \( \dfrac{3}{5}x^{-5/8} = \dfrac{3}{5x^{5/8}} \)
(e) \( (x^{-4})^3 = x^{-12} = \dfrac{1}{x^{12}} \)
(f) \( x^{-5} \times x^{-7} = x^{-5-7} = x^{-12} = \dfrac{1}{x^{12}} \)
(g) \( \left[\left(\dfrac{1}{x}\right)^{-1/2}\right]^{-2/3} = \left(\dfrac{1}{x}\right)^{(-1/2)(-2/3)} = \left(\dfrac{1}{x}\right)^{1/3} = x^{-1/3} = \dfrac{1}{x^{1/3}} \)
(h) \( 3x^{2/5} \times 3x^{-12/5} = 9\,x^{2/5-12/5} = 9\,x^{-10/5} = 9x^{-2} = \dfrac{9}{x^2} \)
(i) \( (3x^2)^0 \times \dfrac{1}{(3x^2)^{-1}} \times 4x^2 \)
\( = 1 \times (3x^2) \times 4x^2 \) [since anything⁰ = 1, and \( \frac{1}{a^{-1}} = a \)]
\( = 12x^4 \)
Evaluate
(a) \( 8^{2/3} \)
\( = (2^3)^{2/3} = 2^{3 \times 2/3} = 2^2 = 4 \)
(b)
\( 27^{-2/3} \)
\( = (3^3)^{-2/3} = 3^{3 \times (-2/3)} = 3^{-2} = \dfrac{1}{9} \)
(c)
\( \dfrac{1}{81^{-3/4}} \)
\( = 81^{3/4} = (3^4)^{3/4} = 3^{4 \times 3/4} = 3^3 = 27 \)
(d)
\( 27^{1/3} \times 16^{-1/4} \)
\( = (3^3)^{1/3} \times (2^4)^{-1/4} = 3^1 \times 2^{-1} = 3 \times \dfrac{1}{2} = \dfrac{3}{2} \)
(e)
\( (125)^{-2/3} \)
\( = (5^3)^{-2/3} = 5^{-2} = \dfrac{1}{25} \)
(f)
\( (256)^{-3/8} \)
\( = (2^8)^{-3/8} = 2^{-3} = \dfrac{1}{8} \)
(g)
\( \left[(625)^{3/4}\right]^{-4/3} \)
\( = (625)^{(3/4)(-4/3)} = (625)^{-1} = \dfrac{1}{625} \)
(h)
\( 25^{3/2} \times 16^{-3/4} \times 5^{-2} \times 1024^{3/2} \)
\( = (5^2)^{3/2} \times (2^4)^{-3/4} \times 5^{-2} \times (2^{10})^{3/2} \)
\( = 5^3 \times 2^{-3} \times 5^{-2} \times 2^{15} \)
\( = 5^{3-2} \times 2^{-3+15} = 5^1 \times 2^{12} = 5 \times 4096 = 20480 \)
(i)
\( \dfrac{1}{\left[(3^4)^{1/2}\right]^{-2}} \)
\( (3^4)^{1/2} = 3^2 \), so \( \left[3^2\right]^{-2} = 3^{-4} \)
\( \dfrac{1}{3^{-4}} = 3^4 = 81 \)
(j)
\( \dfrac{4^3 \times 16^{-1/4} \times 1}{2^6} \times \dfrac{1}{2^{-7}} \)
\( 4^3 = 2^6,\quad 16^{-1/4} = (2^4)^{-1/4} = 2^{-1} \)
\( = \dfrac{2^6 \times 2^{-1}}{2^6} \times 2^7 = 2^{6-1-6+7} = 2^6 = 64 \)
(k)
\( (125)^{-2/3} \times (64)^{4/3} \)
\( = (5^3)^{-2/3} \times (2^6)^{4/3} = 5^{-2} \times 2^8 = \dfrac{256}{25} \)
(l)
\( (32)^{-2/5} \div (125)^{-2/3} \)
\( (32)^{-2/5} = (2^5)^{-2/5} = 2^{-2} = \dfrac{1}{4} \)
\( (125)^{-2/3} = 5^{-2} = \dfrac{1}{25} \)
\( \dfrac{1/4}{1/25} = \dfrac{25}{4} \)
(m)
\( 4 \times 81^{-1/2} \times \left(81^{1/2} + 81^{3/2}\right) \)
\( 81^{1/2} = 9,\quad 81^{3/2} = 9^3 = 729 \)
\( = 4 \times \dfrac{1}{9} \times (9 + 729) = \dfrac{4}{9} \times 738 = \dfrac{2952}{9} = 328 \)
(n)
\( (256)^{-(4-3/2)} = (256)^{-5/2} \)
\( = (2^8)^{-5/2} = 2^{-20} = \dfrac{1}{2^{20}} = \dfrac{1}{1048576} \)
(o)
\( \dfrac{36^{7/2} – 36^{9/2}}{36^{5/2}} \)
\( = 36^{7/2-5/2} – 36^{9/2-5/2} = 36^1 – 36^2 = 36 – 1296 = -1260 \)
(p)
\( \dfrac{(64)^{-1/6} \times (216)^{-1/3} \times (81)^{1/4}}{(512)^{-1/3} \times (16)^{1/4} \times (9)^{-1/2}} \)
\( 64^{-1/6} = (2^6)^{-1/6} = 2^{-1};\quad 216^{-1/3} = (6^3)^{-1/3} = 6^{-1};\quad 81^{1/4} = (3^4)^{1/4} = 3 \)
\( 512^{-1/3} = (8^3)^{-1/3} = 8^{-1};\quad 16^{1/4} = (2^4)^{1/4} = 2;\quad 9^{-1/2} = (3^2)^{-1/2} = 3^{-1} \)
Numerator \( = 2^{-1} \times 6^{-1} \times 3 = \dfrac{3}{12} = \dfrac{1}{4} \)
Denominator \( = 8^{-1} \times 2 \times 3^{-1} = \dfrac{2}{24} = \dfrac{1}{12} \)
\( \dfrac{1/4}{1/12} = 3 \)
(q)
\( \left(\dfrac{256}{6561}\right)^{-5/8} \)
\( 256 = 4^4,\ 6561 = 9^4 \), so \( \dfrac{256}{6561} = \left(\dfrac{4}{9}\right)^4 \)
\( \left[\left(\dfrac{4}{9}\right)^4\right]^{-5/8} = \left(\dfrac{4}{9}\right)^{-5/2} = \left(\dfrac{9}{4}\right)^{5/2} \)
\( = \dfrac{9^{5/2}}{4^{5/2}} = \dfrac{(3^2)^{5/2}}{(2^2)^{5/2}} = \dfrac{3^5}{2^5} = \dfrac{243}{32} \)
(r)
\( \left(\dfrac{117649}{1771561}\right)^{1/6} \)
\( 117649 = 7^6,\quad 1771561 = 11^6 \)
\( \left[\left(\dfrac{7}{11}\right)^6\right]^{1/6} = \dfrac{7}{11} \)
(s)
\( (5^2 + 12^2)^{1/2} \)
\( = (25 + 144)^{1/2} = (169)^{1/2} = 13 \)
(t)
\( (0.04)^{3/2} \)
\( 0.04 = \dfrac{4}{100} = \left(\dfrac{2}{10}\right)^2 \)
\( \left[\left(\dfrac{2}{10}\right)^2\right]^{3/2} = \left(\dfrac{2}{10}\right)^3 = \dfrac{8}{1000} = 0.008 \)
(u)
\( (0.008)^{2/9} \)
\( 0.008 = \dfrac{8}{1000} = \left(\dfrac{2}{10}\right)^3 \)
\( \left[\left(\dfrac{2}{10}\right)^3\right]^{2/9} = \left(\dfrac{2}{10}\right)^{2/3} \)
(v)
\( (0.000064)^{5/6} \)
\( 0.000064 = \dfrac{64}{1000000} = \left(\dfrac{2}{10}\right)^6 \)
\( \left[\left(\dfrac{2}{10}\right)^6\right]^{5/6} = \left(\dfrac{2}{10}\right)^5 = \dfrac{32}{100000} = 0.00032 \)
Simplify: \( \left\{ \sqrt[3]{x^4 y} \times \dfrac{1}{\sqrt[4]{x^2 y^8}} \right\}^{-6} \)
\( \sqrt[3]{x^4 y} = x^{4/3} y^{1/3} \)
\( \sqrt[4]{x^2 y^8} = x^{2/4} y^{8/4} = x^{1/2} y^2 \)
Inside the braces:
\( x^{4/3} y^{1/3} \times x^{-1/2} y^{-2} = x^{4/3 – 1/2}\, y^{1/3 – 2} \)
\( x: \ \dfrac{4}{3}-\dfrac{1}{2} = \dfrac{8-3}{6} = \dfrac{5}{6} \)
\( y: \ \dfrac{1}{3}-2 = \dfrac{1-6}{3} = -\dfrac{5}{3} \)
So inside \( = x^{5/6}\, y^{-5/3} \)
Raise to power \(-6\):
\( \left(x^{5/6}\right)^{-6} \left(y^{-5/3}\right)^{-6} = x^{-5}\, y^{10} = \dfrac{y^{10}}{x^5} \)
Simplify: \( \sqrt[4]{x^4 y^4} \div \sqrt[3]{x^3 y^3} \)
\( \sqrt[4]{x^4 y^4} = (x^4 y^4)^{1/4} = x^{1} y^{1} = xy \)
\( \sqrt[3]{x^3 y^3} = (x^3 y^3)^{1/3} = x^{1} y^{1} = xy \)
\( \dfrac{xy}{xy} = 1 \)
Determine x
(a) \( 3^{1/5} \times 3^{1/6} = 3^{-x} \)
\( 3^{1/5+1/6} = 3^{-x} \)
\( \dfrac{1}{5}+\dfrac{1}{6} = \dfrac{6+5}{30} = \dfrac{11}{30} \)
So \( -x = \dfrac{11}{30} \ \Rightarrow\ x = -\dfrac{11}{30} \)
(b)
\( 800 = 8 \times 10^8 \times x^{-3/2} \)
\( x^{-3/2} = \dfrac{800}{8 \times 10^8} = \dfrac{100}{10^8} = 10^{-6} \)
\( x^{-3/2} = 10^{-6} \)
Raise both sides to power \(-2/3\):
\( x = 10^{-6 \times (-2/3)} = 10^{4} = 10000 \)
By what number should we multiply \( 81^{3/16} \) so that the product becomes \( 3^{5/4} \)?
Let the required number be \(k\).
\( 81^{3/16} \times k = 3^{5/4} \)
\( 81 = 3^4 \ \Rightarrow\ 81^{3/16} = 3^{4 \times 3/16} = 3^{3/4} \)
\( 3^{3/4} \times k = 3^{5/4} \)
\( k = 3^{5/4 – 3/4} = 3^{2/4} = 3^{1/2} = \sqrt{3} \)
By what number should we divide \( (121)^{7/8} \) to obtain \( (1331)^{3/4} \)?
Let the required number be \(k\).
\( \dfrac{(121)^{7/8}}{k} = (1331)^{3/4} \)
\( 121 = 11^2,\quad 1331 = 11^3 \)
\( (121)^{7/8} = 11^{2 \times 7/8} = 11^{7/4} \)
\( (1331)^{3/4} = 11^{3 \times 3/4} = 11^{9/4} \)
\( k = \dfrac{11^{7/4}}{11^{9/4}} = 11^{7/4-9/4} = 11^{-2/4} = 11^{-1/2} = \dfrac{1}{\sqrt{11}} \)
If \( x = 24 \), find the value of \( (2x)^x \)
Note: \(x=24\) makes the answer astronomically large ( \((48)^{24}\) ), which is unusual for this exercise level. This is very likely a typo for \(x = 2/4 = \tfrac{1}{2}\) or similar in the original text — the worked answer below is for the value exactly as printed.
\( (2x)^x = (2 \times 24)^{24} = (48)^{24} \)
This is the exact simplified form. As a numeral, \(48^{24}\) is a 41-digit number — please double-check the value of \(x\) against your textbook if a smaller numeric answer was expected.
Determine x and y so that \( 3^{x+y} = 81 \) and \( 3^{x-y} = 3 \)
From the first equation: \( 3^{x+y} = 81 = 3^4 \ \Rightarrow\ x + y = 4 \)
From the second equation: \( 3^{x-y} = 3^1 \ \Rightarrow\ x – y = 1 \)
Adding the two equations:
\( (x+y)+(x-y) = 4 + 1 \ \Rightarrow\ 2x = 5 \ \Rightarrow\ x = \dfrac{5}{2} \)
Subtracting:
\( (x+y)-(x-y) = 4 – 1 \ \Rightarrow\ 2y = 3 \ \Rightarrow\ y = \dfrac{3}{2} \)
So \( x = \dfrac{5}{2},\quad y = \dfrac{3}{2} \)