9类RD Sharma解决方案–第2章实数指数-练习2.2 |套装2

问题12确定(8倍)^的x，如果^{9×+ 2} = 240 + ^9×。

解决方案：

We have,

=> 9^x+2 = 240 + 9^x

=> 9^x+2 − 9^x = 240

=> 9^x (9² − 1) = 240

=> 9^x = 240/80

=> 3^2x = 3

=> 2x = 1

=> x = 1/2

Therefore, (8x)^x = [8 × (1/2)]^1/2

= 4^1/2

= 2

问题13.如果3 ^{x + 1} = 9 ^x−2 ，则找到2 ^{1 + x}的值。

解决方案：

We have,

=> 3^x+1 = 9^x−2

=> 3^x+1 = (3²)^x−2

=> 3^x+1 = 3^2x−4

=> x + 1 = 2x − 4

=> x = 5

Therefore, 2^1+x = 2¹⁺⁵

= 2⁶ 

= 64

问题14.如果3 ^4x =(81) ^-1和(10) ^{1 / y} = 0.0001，则找到2 ^{-x + 4y}的值。

解决方案：

We are given,

=> 3^4x = (81)⁻¹

=> 3^4x = (3⁴)⁻¹

=> 3^4x = (3)⁻⁴

=> 4x = −4

=> x = −1

And also, (10)^1/y = 0.0001

=> (10)^1/y = (10)⁻⁴

=> 1/y = −4

=> y = −1/4

Therefore, 2^−x+4y = 2^1+4(−1/4) 

= 2¹⁻¹ 

= 1

问题15：如果5 ^3x = 125和10 ^y = 0.001。找出x和y。

解决方案：

We are given,

=> 5^3x = 125

=> 5^3x = 5³

 => 3x = 3

=> x =1

Also, (10)^y = 0.001

=> 10^y = 10⁻³

=> y = −3

Therefore, the value of x is 1 and the value of y is –3.

问题16：解下列方程式：

(i)3 ^{x + 1} = 27×3 ⁴

解决方案：

We have,

=>  3^x+1 = 27 × 3⁴

=> 3^x+1 = 3³ × 3⁴

=> 3^x+1 = 3⁷

=> x + 1 = 7

=> x = 6

(ii) $4^{2x}=\left(\sqrt[3]{16}\right)^{\frac{-6}{y}}=(\sqrt{8})^2$ 

解决方案：

We have,

=> $4^{2x}=\left(\sqrt[3]{16}\right)^{\frac{-6}{y}}=(\sqrt{8})^2$

=> $(2^2)^{2x}=\left(\sqrt[3]{2^4}\right)^{\frac{-6}{y}}=(\sqrt{2^3})^2$

=> $(2)^{4x}=\left(2^\frac{4}{3}\right)^{\frac{-6}{y}}=(2^\frac{3}{2})^2$

=> $(2)^{4x}=(2)^{\frac{-8}{y}}=2^3$

=> 4x = −8/y = 3

=> x = 3/4 and y = −8/3

(iii)3 ^x-1 ×5 ^2y-3 = 225

解决方案：

We have,

=> 3^x−1 × 5^2y−3 = 225

=> 3^x−1 × 5^2y−3 = 3² × 5²

=> x − 1 = 2 and 2y − 3 = 2 

=> x = 3 and 2y = 5

=> x = 3 and y = 5/2

(iv)8 ^{x + 1} = 16 ^{y + 2}和(1/2) ^{3 + x} =(1/4) ^3y

解决方案：

We have,

=> 8^x+1 = 16^y+2

=> (2³)^x+1 = (2⁴)^y+2

=> 2^3x+3 = 2^4y+8

=> 3x + 3 = 4y + 8  . . . . (1)

Also, (1/2)^3+x = (1/4)^3y

=> (1/2)^3+x = [(1/2)²]^3y

=> (1/2)^3+x = (1/2)^6y

=> 3 + x = 6y

=> x = 6y − 3   . . . . (2)

Putting (2) in (1), we get,

=> 3(6y − 3) + 3 = 4y + 8

=> 18y − 9 + 3 = 4y + 8

=> 14y = 14

=> y = 1

Putting y = 1 in (2), we get,

x = 6(1) − 3 = 6 − 3 = 3

Therefore, the value of x is 1 and the value of y is –3.

(v)4 ^x-1 ×(0.5) ^3-2x =(1/8) ^x

解决方案：

We have,

=> 4^x−1 × (0.5)^3−2x = (1/8)^x

=> (2²)^x−1 × (1/2)^3−2x = [(1/2)³]^x

=> 2^2x−2 × 2^2x−3 = 2^−3x

=> 2^{2x−2+2x−3} = 2^−3x

=> 2^4x−5 = 2^−3x

=> 4x − 5 = −3x

=> 7x = 5

=> x = 5/7

(六) $\sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2x}$ 

解决方案：

We have,

=> $\sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2x}$

=> $\left(\frac{a}{b}\right)^{\frac{1}{2}}=\left(\frac{a}{b}\right)^{2x-1}$

=> 1/2 = 2x − 1

=> 2x = 3/2

=> x = 3/4

问题：17.如果a和b是不同的正质数，则$\sqrt[3]{a^6b^{-4}}=a^xb^{2y}$找出x和y。

解决方案：

We have,

=> $\sqrt[3]{a^6b^{-4}}=a^xb^{2y}$

=> (a⁶ b⁻⁴)^1/3 = a^xb^2y

=> a^6/3 b^−4/3 = a^xb^2y

=> a² b^−4/3 = a^xb^2y

=> x = 2 and 2y = −4/3

=> x = 2 and y = −2/3

问题18：如果a和b是不同的正质数，则，

(一世) $\left(\frac{a^{-1}b^{2}}{a^2b^{-4}}\right)^7÷\frac{a^{3}b^{-5}}{a^{-2}b^{3}}=a^xb^y$ ，找到x和y。

解决方案：

We have,

=> $\left(\frac{a^{-1}b^{2}}{a^2b^{-4}}\right)^7÷\frac{a^{3}b^{-5}}{a^{-2}b^{3}}=a^xb^y$ 

=> (a⁻¹⁻² b²⁺⁴)⁷ ÷ (a³⁺² b⁻⁵⁻³) = a^xb^y

=> (a⁻³ b⁶)⁷ ÷ (a⁵ b⁻⁸) = a^xb^y

=> (a⁻²¹ b⁴²) ÷ (a⁵ b⁻⁸) = a^xb^y

=> (a⁻²¹⁻⁵ b⁴²⁺⁸) = a^xb^y

=> (a⁻²⁶ b⁵⁰) = a^xb^y

=> x = −26, y = 50

(ii)(a + b) ^-1 (a ^-1 + b ^-1 )= a ^x b ^y ，求出x + y + 2。

解决方案：

We have,

=> (a + b)⁻¹(a⁻¹ + b⁻¹) = a^xb^y

=> $(\frac{1}{a+b})(\frac{1}{a}+\frac{1}{b})$ = a^xb^y 

=> $(\frac{1}{a+b})(\frac{b+a}{ab})$ = a^xb^y 

=> 1/ab = a^xb^y

=> a⁻¹b⁻¹ = a^xb^y

=> x = −1 and y = −1

So, x+y+2 = −1−1+2 = 0.

问题19.如果2 ^x ×3 ^y ×5 ^z = 2160，则求出x，y和z。因此，计算3 ^x ×2 ^-y ×5 ^-z的值。

解决方案：

We are given,

=> 2^x × 3^y × 5^z = 2160

=> 2^x × 3^y × 5^z = 2⁴ × 3³ × 5¹

=> x = 4, y = 3, z = 1

Therefore, 3^x × 2^−y × 5^−z = 3⁴ × 2⁻³ × 5⁻¹

= (81) (1/8) (1/5)

= 81/40

问题20.如果1176 = 2 ^a ×3 ^b ×7 ^c ，请找到a，b和c的值。因此，将2 ^a ×3 ^b ×7 ^-c的值作为分数计算。

解决方案：

We are given,

=> 1176 = 2^a × 3^b × 7^c

=> 2³ × 3¹ × 7² = 2a × 3b × 7c

=> a = 3, b = 1, c = 2 

Therefore, 2^a × 3^b × 7^−c = 2³ × 3¹ × 7⁻²

= (8) (3) (1/49)

= 24/49

问题21.简化

(一世) $\left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c}\left(\frac{x^{c+a}}{x^b}\right)^{c-a}$

解决方案：

We have,

= $\left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c}\left(\frac{x^{c+a}}{x^b}\right)^{c-a}$

= (x^a+b−c)^a−b (x^b+c−a)^b−c (x^c+a−b)^c−a

= $(x^{a^2-b^2-ca+cb})(x^{b^2-c^2-ab+ac})(x^{c^2-a^2-bc+ab})$

= $(x^{a^2-b^2-ca+cb+b^2-c^2-ab+ac+c^2-a^2-bc+ab})$

= x⁰

= 1

(ii) $\sqrt[lm]{\frac{x^l}{x^m}}×\sqrt[mn]{\frac{x^m}{x^n}}×\sqrt[nl]{\frac{x^n}{x^l}}$

解决方案：

We have,

=> $\sqrt[lm]{\frac{x^l}{x^m}}×\sqrt[mn]{\frac{x^m}{x^n}}×\sqrt[nl]{\frac{x^n}{x^l}}$ 

=> $(x^{l-m})^{\frac{1}{lm}}×(x^{m-n})^{\frac{1}{mn}}×(x^{n-l})^{\frac{1}{nl}}$

=> $x^{\frac{l-m}{lm}}×x^{\frac{m-n}{mn}}×x^{\frac{n-l}{nl}}$

=> $x^{\frac{l-m}{lm}+\frac{m-n}{mn}+\frac{n-l}{nl}}$

=> $x^{\frac{nl-mn+ml-nl+mn-ml}{mnl}}$

=> $x^{\frac{0}{mnl}}$

=> x⁰

= 1

问题22：表明$\frac{(a+\frac{1}{b})^m×(a-\frac{1}{b})^n}{(b+\frac{1}{a})^m×(b-\frac{1}{a})^n}=(\frac{a}{b})^{m+n}$ 。

解决方案：

We have,

L.H.S. = $\frac{(a+\frac{1}{b})^m×(a-\frac{1}{b})^n}{(b+\frac{1}{a})^m×(b-\frac{1}{a})^n}$

= $\frac{(\frac{ab+1}{b})^m×(\frac{ab-1}{b})^n}{(\frac{ab+1}{a})^m×(\frac{ab-1}{a})^n}$

= $(\frac{a}{b})^m×(\frac{a}{b})^n$

= $(\frac{a}{b})^{m+n}$

= R.H.S.

Hence proved.

问题23(i)如果a = x ^{m + n} y ^l ，b = x ^{n + l} y ^m和c = x ^{l + m} y ⁿ ，则证明a ^-n b ^n-l c ^l-m = 1 。

解决方案：

Given, a = x^m+ny^l, b = x^n+ly^m and c = x^l+myⁿ. 

We have,

L.H.S. = a^m−n b^n−l c^l−m

= (x^m+ny^l)^m−n(x^n+ly^m)^n−l(x^l+myⁿ)^l−m

= $(x^{m^2-n^2}y^{lm-ln})(x^{n^2-l^2}y^{mn-ml})(x^{l^2-m^2}y^{nl-mn})$

= $x^{m^2-n^2+n^2-l^2+l^2-m^2}y^{lm-ln+mn-ml+nl-mn}$

= x⁰y⁰

= 1

= R.H.S.

Hence proved.

(ii)如果x = a ^{m + n} ，y = a ^{n + 1}和z = a ^{l + m} ，则证明x ^m y ⁿ z ^l = x ⁿ y ^l z ^m 。

解决方案：

Given, x = a^m+n, y = a^n+l and z = a^l+m.

We have,

L.H.S. = x^myⁿz^l

= (a^m+n)^m (a^n+l)ⁿ (a^l+m)^l

= $a^{m^2+mn+n^2+ln+l^2+ml}$

= $a^{mn+n^2}×a^{nl+l^2}×a^{lm+m^2}$

= (a^m+n)ⁿ (a^n+l)^l (a^l+m)^m

= xⁿy^lz^m

= R.H.S.

Hence proved.