Exercise 2.4 (Solutions)

Question 1

Use law of exponent to simplify.

• (i) (243)−23(32)−15(√196)−1
• (ii) (2x5y−4)(−8x−3y2)
• (iii) (x−2y−1z−4x4y−3z0)−3
• (iv) (81)n.35−(3)4n−1(243)(92n)(33)

Solution

(i)

(243)−23(32)−15(√196)−1=(35)−23(25)−15(142)−1×12=3−1032−114−1=3−1032−12−1×7−1=3−1037−1=73103=73(9)×13×313=733.3√3=727.3√3

(ii)

(2x5y−4)(−8x−3y2)=(2)(−8)x5.y−4.x−3y2=−16x5−3y−4+2=−16x2y−2=−16x2y2

(iii)

(x−2y−1z−4x4y−3z0)−3=(y−1+3x4+2z4)−3=(y2x6z4)−3=y2(−3)x6(−3)z4(−3)=y−6x−18z−12=x18z12y6

(iv)

(81)n.35−(3)4n−1(243)(92n)(33)=(34)n.35−34n−1(3)5(32)2n.33=34n+5−34n−1+534n.33=34n+4.31−34n+434n+3=34n+4(3−1)34n+3=34n+4−4n−3×2=3×2=6

Question 2

Show that

(xaxb)a+b×(xbxc)b+c×(xcxa)c+a=1

Solution

(xa−b)a+b×(xb−c)b+c×(xc−a)c+a=xa2−b2×xb2−c2×xc2−a2=xa2−b2+b2−c2+c2−a2=x0=1

Question 3

• Simplify
• (i) 21/3(27)1/3(60)1/2(180)1/2(4)−1/391/4
• (ii) 216232512.04−12−−−−−−−√
• (iii) 523÷(52)3
• (iv) (x3)2÷x32

Soluton

(i)

21/3(27)1/3(60)1/2(180)1/2(4)−1/391/4=21/3(33)1/3(22.3.5)1/2(22.32.5)1/2(22)−1/3(32)1/4=21/3(33)1/3(22.3.5)1/2(22.32.5)1/2(22)−1/3(32)1/4=21/33.2.31/2.51/22.3.51/22−2/331/2=21/3+2/3=21+2/3=23/3=2

(ii)

2162/3251/2.04−1/2−−−−−−−−−−√=632/3521/24100−1/2−−−−−−−−−⎷=625125−1/2−−−−−−⎷=62.5(152)−1/2−−−−−−−−⎷=62.55−−−−√=621/2=6

(iii)

523÷(52)3=58÷56=5856=58−6=52=25

(iv)

(x3)2÷x32=x6÷x9=x6x9=1x6−9=1x−3=1x3