Exercise 6 C
Question 1
(i) 50×4−1+813
(ii) 3√(64)−4(125)−2
(iii) (9−3×163/2)1/6
(iv) 3√(16)−3/4×(125)−2
(v) (32)−2/5÷(216)−2/3
Sol :
(i) 50×4−1+813
=1×14+(2×2×2)13
=14+(23)1/3=14+(23×13)
=14+21
=2+14
⇒214
(ii) 3√(64)−4(125)−2
=[(64)−4(125)−2]1/3
=[(64)−4×1/3]×[(125)−2×1/3]
=[(64)−4/3]×(125)−2/3
=(43)−4/3×(53)−2/3
=43×(−4/3)×53×(−2/3)
=4−4×5−2
=144×52=14×4×4×4×5×5
=16400
(iii) (9−3×163/2)1/6
=[{13)2]−3×(24)3/2]1/6
=[(3)−6×(2)6]1/6
=3−6×1/6×26×1/6
=3−1×21=13×2
=23
(iv) 3√(16)−3/4×(125)−2
=3√(24)−3/4×(53)−2
=[24×(−3/4)×53×(−2)]1/3
=[2−3×5−6]1/3
[123×56]1/3
1(23)1/2×(56)1/3=121×52
=12×5×5=150
(v) (32)−2/5÷(216)−2/3
=(25)−2/5÷(63)−2/3
=(2)−2/5×5+(6)3×(−2/3)
=2−2÷6−2
=(26)−2=(62)2=6×62×2
= 364
=9
Question 2
(i) (4125)−2/3
(ii) 93/2−3×(5)0−(181)−1/2
(iii) (14)−2−3 (8)2/3 х 40+(316)−1/2
(iv) 163/4+2(12)−1×30
(v) (81)3/4−(132)−2/5+(8)1/3(12)−2(2)0
Sol:
(i) (4125)−2/3
1(265625)1/4+(√253√64)0
=(453)−2/3÷1(4454)1/2 +[(52)1/2(43)1/3]0
(45)3×(−2/3)×(45)4×1/4+1
=(45)−2×(45)1+1
=(54)2×(54)−1+1
=(54)2−1+1
=(54)+1
=54+1=114+1
=214 Ans
(ii) 93/2−3×(5)0−(181)−1/2
=(32)3/2−3×1−(134)−1/2
=32×3/2−3−1(34)−1/2
=33−3−1(3)−2
=27−3−(3)2=27−3−9
=27-12
=15 Ans
(iii) (14)−2−3 (8)2/3 х 40+(316)−1/2
=(122)−2−3(23)2/3×1+[(3)2(4)2]−1/2
=1(2)2(−2) −3(23×2/3)× 1+ (34)2^(−1/2)
= 12−4 −3×22×1+(34)−1
=24−3×4+43=16−12+43
=4+113=513
(iv) 163/4+2(12)−1×3
=(24)3/4+2×21×1=23+2×2×1
=2×2×2+2×2=8+4=12 Ans
(v) (81)3/4−(132)−2/5+(8)1/3(12)−2(2)0
=(34)3/4−125(−2/5)+(23×1/3)×1(2)−2×1
=(3)3−1(2)−2+22×221
=27−(2)1+2×22
=27-4+8
=35-4
=31 Ans
Question 3
(a) Evaluate : x1/2×y−1×z2/3 , when x = 9, y = 2, and z = 8.
(b) Evaluate as a fraction :
(278)2/3−(14)−2+(5)0
(c) Evaluate as a fraction :
√14+(0.01)−1/2−(27)2/3
Sol:
(a) x1/2×y−1×z2/3
x=9
y=2
z=8
⇒(9)1/2×(2)−1×(8)2/3
=(32)1/2×12×(23)2/3
=32×1/2×12×23×2/3
=31×12×22=3×12×4
=3×2=6 Ans
(b) (278)2/3−(14)−2+(5)0
=(3323)2/3−(122)−2+1
33×2/323×2/3−121×(−2)+1
=3222−12−4+1=94−24+1
=94−16+1=214−16+1
=−1234 Ans.
(c) √14+(0.01)−1/2−(27)2/3
= (122)1/2+(0.1)2×(−1/2)−(33)2/3
=122×1/2+(0.1)−1−32
=12+1(0.1)−9=12+101−9
=1012−9=112 Ans
Question 4
(i) (64)5n/6(27)−n/6(12)−n/2⋅26n
(ii) 6⋅(8)n+1+1623n⋅210⋅23n+1−7⋅(8)n
(iii) 42n⋅2n+12n−3⋅42n+1
(iv) (5)2n+3−(25)n+2((125)n+1]2/3
Sol:
(i) (64)5n/6(27)−n/6(12)−n/2⋅26n
=(26)5n/6(39)−n/6(22×3)−n/2⋅26n
= (2)6×5n6⋅(3)3×(−n6)(2)−n2×2⋅(3)−n2⋅26n
=25n⋅3−n/22−n⋅3−n/2⋅26n=(2)5n+6n+n⋅(3)−n2+n2
==212n⋅30=(12)12n⋅1.
=(2)12n
(ii) 6⋅(8)n+1+1623n⋅210⋅23n+1−7⋅(8)n
=6⋅(23)n+1+16⋅23n−210⋅23n+1−7⋅(23)n
= 6⋅23n+3+16⋅23n−210⋅23n+1−7⋅23n
=6⋅23n⋅23+16⋅23n⋅2−210⋅23n⋅2−(7⋅2)3n
=23n(6×23+16×2−2)23n(10×2−7)
=6×8+16×1420−7
=48+413=5213=4
(iii) 42n⋅2n+12n−3⋅42n+1
=(22)2n⋅2n+12(n−3)(22)2n−1
==24n⋅2n+12n−3⋅24n+2=24n+n+12n−3+4n+2
=25n+125n−1
=25n+1−5n+1
=22
=4
(iv) (5)2n+3−(25)n+2((125)n+1]2/3
=53n+3−(5)m+2[(53)n+3]2/3
=52n⋅53⋅52n⋅545(3n+5)3/2
⇒52n⋅53−52n⋅5452n⋅52
=52n(53−54)52n⋅52
=125−62525
=−50025
=-20 Ans
Question 5
If 49n+1⋅7n−(343)n73m⋅2n=3343 , prove that m = (m + 1)
Sol:
49n+1⋅7n−(343)n73m⋅2n⇒3343⇒(72)n+1⋅7n−(73)n73m⋅2n=3343⇒7n+2⋅7n−73n73m⋅24−3343⇒73n(72−1)16⋅73m=3343⇒73n×4816⋅73m=3343
=73n73m=3×16343×48=1343
=(7n)3(7m)3=173=(17)3⇒7n7m=177.7n=7m
⇒7n+1=7m (comparing both sides) ⇒m=n+1
Hence proved.
Question 6
If 2205=3a×5b×7c
(i) Find the numerical values of a, b and c
(ii) Hence evaluate 3a×5−b×7−c
Sol:
(i) 2205=3a×5b×7c
3220537355245749771
On comparing = a=2
b=1
c=2
so 32×51×72=3a×5b×7c
(ii)
3a×5−b×7−c=3a5b×7c=3251×72=95×49⇒92459245 Ans.
Question 7
If [b3c−2b−4c3]−3+[b−1cb2.c−2]5=bx.cx , prove that x + y + 6 = 0
Sol:
[b3c−2b−4c3]−3÷[b−1cb2⋅c−2]5=bx⋅cy
Now R.H.S. =x+y+6=−6+0+6=0
Hence proved.
Question 8
Prove that
(i) (x+y)−1(x−1+y−1)=1xy
(ii) (x−1+y−1)−1=xyx+y
(iii) a+b+ca−1b−1+b−2c−1+c−1a−1=abc
(iv) 11+xb−a+xc−a+11+xa−b+xc−b+11+xb−c+xa−c=1
(v) ab√xaxb⋅bc√xbxc⋅ca√x2xa=1
Sol:
(i)
(x+y)−1(x−1+y−1)=1xy L.H.S. =(x+y)−1(x−1+y−2)=1x+y×(1x+1y)=1x+y×y+5xy
=1x+y×x+yxy⇒1xy
Hence pruerd,
L.H.S = R.H.S.
(ii) (x−1+y−1)−1=xyx+y
L.H.S
(x−1+y−1)−1=[1x+1y]−1=[y+xxy]−1=xyx+y R.H.S.
(iii) a+b+ca−1b−1+b−2c−1+c−1a−1=abc
L.H.S
a+b+ca−1b−1+b−1c−1+c−1a−1=a+b+c1a1b+1b+1c+1c1a
=a+b+cc+a+babc⇒(a+b+c)×abc(a+b+c)
=abc R.H.S (Hence proved)
(iv) 11+xb−a+xc−a+11+xa−b+xc−b+11+xb−c+xa−c=1
L.H.S
=1xa−a+xb−a⋅xc−a+1xb−b+xa−b+xc−b+1xc−c+xb−c+xa−c
=1x−a(xa+xb+xc)+1x−b(xb+xa+xc)+1x−c(x2+xa+xb)
=1(xa+xb+xc)[1x−a+1x−b+1x−c]
1(xa+−xb+x2)[xa+xb+xc1]=1
L.H.S =R.H.S PROVED
(v) ab√xaxb⋅bc√xbxc⋅ca√x2xa=1
xa−bab⋅xb−cbc⋅xc−aca
=xa−bab+b−cbc+c−aca
xabc−bc+ab−ac+bc−ababc
xDabc
=x0=1L⋅H⋅S=R⋅H⋅S
Hence, proved
Question 9
(i) (p1/3−p−1/3)(p2/3+1+p−2/3)
(ii) (√11+√3)1/3(√11−√3)1/3
Sol:
(i)(p1/3−p−1/3)(p2/3+1+p−2/3)
if p1/3=x and p−1/3=y
=(a2−y2)(x2+xy+y2)
=x3−y3=(p1/3)3−(p−1/3)3 =p1−p−1=p−1p Ans.
(ii) (√11+√3)1/3(√11−√3)1/3
if √11=x
√3=y
=(x+y)1/3(x−y)1/3
[ from formula. ]= (x2−y2)1/3
=[(√11)2−(√3)2]1/3
=(11−3)1/3=(8)1/3
=(23)1/3=23/13
=22=2 Ans
Question 10
If 3x=5y=45z , prove that x=2yzy−z
Sol:
L.H.S
Suppose 3x=5y=45z=a
so, a1/x=3,a1/y=5,1a2=4545=3×3×5⇒32×5=45=a1z=a2x×a1/y=a1z=a2x+1y
In comparing
12=2x+1y=2x=1z−1y ⇒2x=y−zyz ⇒1x=y−z2yz ⇒x=2yzy−z R.H.S.
Hence proved
Question 11
If x=abp−1,y=abq−1 and z=abr−1 prove that xp−rxyr−p.zp−q=1.
Sol;
L.H.S.
x=abp−1y=abq−1z=abr−1
So =x−r−r⋅yr−p⋅zp−q
=(abp−1)q−˙r⋅(ab2−1)r−p⋅(abr−1)p−q
=aq−r⋅b(2−r)(p−1)⋅ar−p⋅b(r−p)(r−1)⋅ap−2⋅b(p−2)(r−1)
=a∘⋅b∘
=1×1=1 R.H.S
Hence proved
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