问题1(i)。简化3(a 4 b 3 ) 10 ×5(a 2 b 2 ) 3
解决方案:
Given 3(a4b3)10 × 5(a2b2)3
= 3 × a40 × b30 × 5 × a6 × b6
= 3 × a46 × b36 × 5 [am × an = am+n]
= 15 × a46 × b36
= 15a46b36
Thus, 3(a4b3)10 × 5(a2b2)3 = 15a46b36
问题1(ii)。简化(2x -2 y 3 ) 3
解决方案:
Given (2x-2y3)3
= 23 × x-6 × y9
= 8 × x-6 × y9 [am × an = am+n]
= 8x-6y9
Thus, (2x-2y3)3 = 8x-6y9
问题1(iii)。简化
解决方案:
Given
=
= [am × an = am+n]
=
= 3/102
= 3/100
Thus,
问题1(iv)。简化
解决方案:
Given
=
=
= [am × an = am+n]
= -2×a2×b5×a-2×b-2
= -2×a2+(-2)×b5+(-2) [am × an = am+n]
= -2×a0×b3
= -2b3 [a0=1]
Thus, =-2b3
问题1(v)。简化
解决方案:
Given
=
= [am × an = am+n]
Thus,
问题1(vi)。简化
解决方案:
Given
= [(am)n = amn]
=
= a18n-54 × a-(2n-4) [am × an = am+n]
= a18n-54-2n+4
= a16n-50
Thus, = a16n-50
问题2(i)如果a = 3且b = -2,则求a a + b b的值
解决方案:
Given a = 3 and b = -2
On substituting the value of a and b in aa + bb, we get
aa + bb = 33 + (-2)-2
= 27 + 1/4
= (108 + 1)/4
= 109/4
Thus, aa + bb = 109/4
问题2(ii)。如果a = 3且b = -2,则找到a b + b a的值
解决方案:
Given a = 3 and b = -2
On substituting the value of a and b in ab + ba, we get
ab + ba = 3-2 + (-2)3
= 1/9 + (-8)
= (1 – 72)/9
= -71/9
Thus, ab + ba = -71/9
问题2(iii)。如果a = 3且b = -2,则找到(a + b) ab的值。
解决方案:
Given a = 3 and b = -2
On substituting the value of a and b in (a + b)ab, we get
(a + b)ab = (3 + (-2))3×-2
= (1)-6
= 1
Thus, (a + b)ab = 1
问题3(i)。证明
解决方案:
Let us first solve left-hand side of the given equation
By using the formula (am)n = amn, we get
=
By using the formula am/an = am-n, we get
=
=
=
By using the formula am × an = am+n , we get
=
= x
= 1
= Right-hand side of the given equation
Thus, we proved that
问题3(ii)。证明
解决方案:
Let us consider the left-hand side of the given equation
By using the formula, (am)n = amn, we get
=
=
= [am × an = am+n]
= 1
= Right-hand side of the given equation
Thus, we proved that
问题3(iii)。证明
解决方案:
Let us first solve left-hand side of the given equation
By using the formula (am)n = amn, we get
=
By using the formula am/an = am-n, we get
=
=
=
By using the formula am × an = am+n , we get
=
=
= Right-hand side of the given equation
Thus, we proved that
问题4(i)。证明
解决方案:
Let us first consider the left-hand side of given equation
=
=
=
=
= 1
= Right-hand side of the given equation
Thus, we proved that
问题4(ii)。证明
解决方案:
Let us first consider the left-hand side of given equation
=
=
=
=
= 1
= Right-hand side of the given equation
Thus, we proved that
问题5(i)。证明
解决方案:
Let us first consider the left-hand side of given equation
=
=
= abc
= Right hand side of the given equation
Thus, we proved that
问题5(ii)。证明
解决方案:
Let us first consider the left hand side of given equation
=
=
=
=
= Right hand side of the given equation
Thus, we proved that
问题6。如果abc = 1,则表明
解决方案:
Given abc = 1
⇒ c = 1/ab
Let us first consider the left-hand side of given equation
=
=
=
By substituting the value of c in above equation, we get
=
=
=
=
=
= 1
= Right hand side of the given equation
Thus, we have shown that if abc = 1,
问题7(i)。简化
解决方案:
Given
=
=
= [am × an = am+n]
=
= 33n+2-(3n-3) [am/an = am-n]
= 35
= 243
Thus, = 243
问题7(ii)。简化
解决方案:
Given
=
=
= [am × an = am+n]
=
= 4/24
= 1/6
Thus, = 1/6
问题7(iii)。简化
解决方案:
Given,
=
=
= (19 × 5)/5
= 19
Thus,
问题7(iv)。简化
解决方案:
Given
=
=
=
=
=
= (48 + 4)/13
= 52/13
= 4
Thus,
问题8(i)。对x求解方程7 2x + 3 = 1。
解决方案:
Given equation 72x+3 = 1
We know that, for any a∈ Real numbers, a0 = 1
Let a = 7
⇒ 72x+3 = 70
Since the bases are equal, let us equate the exponents
⇒ 2x + 3 = 0
⇒ x = -3/2
Thus, the value of x is -3/2
问题8(ii)。对x求解方程2 x + 1 = 4 x-3 。
解决方案:
Given 2x+1 = 4x-3
We can write 4 = 22
⇒ 2x+1 = 22(x-3)
⇒ 2x+1 = 22x-6
Since the bases are equal, let us equate the exponents
⇒ x + 1 = 2x – 6
⇒ x = 7
Thus, the value of x is 7
问题8(iii)。对x求解方程2 5x + 3 = 8 x + 3。
解决方案:
Given 25x+3 = 8x+3
We know that 8 = 23
⇒ 25x+3 = 23(x+3)
⇒ 25x+3 = 23x+9
Since the bases are equal, let us equate the exponents
⇒ 5x + 3 = 3x + 9
⇒ 5x – 3x = 9 – 3
⇒ 2x = 6
⇒ x = 3
Thus, the value of x is 3
问题8(iv)。对x求解方程4 2x = 1/32。
解决方案:
Given 42x = 1/32
⇒ 22(2x) = 1/32
⇒ 22(2x) × 32 = 1
⇒ 24x × 25 = 1
⇒ 24x+5 = 20
Since the bases are equal, let us equate the exponents
⇒ 4x + 5 = 0
⇒ x = -5/4
Thus, the value of x is -5/4
问题8(v)。求解方程4× –对于x 1×(0.5)3-2x =(1/8)X。
解决方案:
Given 4x – 1 × (0.5)3-2x = (1/8)x
⇒
⇒
⇒ 22(x-1) × 2-(3-2x) = 2-3x
⇒ 22x-2-3+2x = 2-3x
⇒ 24x-5 = 2-3x
Since the bases are equal, let us equate the exponents
⇒ 4x – 5 = -3x
⇒ 7x = 5
⇒ x = 5/7
Thus, the value of x is 5/7
问题8(vi)。对x求解方程2 3x-7 = 256。
解决方案:
Given 23x-7 = 256
⇒ 23x-7 = 28
Since the bases are equal, let us equate the exponents
⇒ 3x – 7 = 8
⇒ x = 15/3
⇒ x = 5
Thus, the value of x is 5
问题9(i)。对x求解方程2 2x – 2 x + 3 + 2 4 = 0。
解决方案:
Given 22x – 2x+3 + 24 = 0
⇒ (2x)2 – 2 × 2x × 22 + (22)2 = 0
⇒ (2x – 22)2 = 0
⇒ 2x – 22 = 0
⇒ 2x = 22
Since the bases are equal, let us equate the exponents
⇒ x = 2
Thus, the value of x is 2
问题9(ii)。对x求解方程3 2x + 4 +1 = 2.3 x + 2。
解决方案:
Given 32x+4 + 1 = 2.3x+2
⇒
⇒
⇒ (3x+2 – 1)2 = 0
⇒ 3x+2 – 1 = 0
⇒ 3x+2 = 30
Since the bases are equal, let us equate the exponents
⇒ x + 2 = 0
⇒ x = -2
Thus, the value of x is -2
问题10.如果49392 = a 4 b 2 c 3 ,则找到a,b和c的值,其中a,b和c是不同的正质数。
解决方案:
Let us first find out prime factorization of 49392
Thus, 49392 = 24 × 32 × 73
Where 2, 3 and 7 are positive primes
49392 = 243273 = a4b2c3
Thus, on comparing, we get
a = 2,b = 3 and c = 7
Thus, the values of a, b and c are 2, 3, 7 respectively.
问题11.如果1176 = 2 a 3 b 7 c ,则找到a,b和c。
解决方案:
Given 1176 = 2a3b7c
Let us first find out prime factorization of 1176
Thus, 1176 = 23 × 31 × 72
1176 = 233172 = 2a3b7c
Thus, on comparing, we get
a = 3, b = 1, c = 2
Thus, the values of a, b and c are 3, 1, 2 respectively.
问题12.给定4725 = 3 a 5 b 7 c ,求
(i)a,b和c的积分值
(ii)2 -a 3 b 7 c的值
解决方案:
Given 4725 = 3a5b7c
(i) Let us first find out prime factorization of 4725
Thus, 4725 = 33 × 52 × 71
4725 = 335271 = 3a5b7c
Thus, on comparing, we get
a = 3,b = 2,c = 1
Thus, the values of a, b and c are 3,2,1 respectively.
(ii) Here a = 3, b = 2, c = 1
On substituting these values in 2-a3b7c
2-a3b7c= 2-3×32×71
= 1/8 × 9 × 7 = 63/8
Thus, the value of 2-a3b7c is 63/8
问题13.如果a = xy p-1 ,b = xy q-1 ,c = xy r-1 ,则证明a qr b rp c pq = 1。
解决方案:
Given a = xyp-1, b = xyq-1, c = xyr-1
aq-rbr-pcp-q=
=
= xq-r+r-p+p-q y(p-1)(q-r)+(r-p)(q-1)+(p-q)(r-1)
= xq-r+r-p+p-q ypq-q-pr+r+rq-r-pq+p+pr-p-qr+q
= x0y0
= 1
Thus, we proved that aq-rbr-pcp-q = 1