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Class 9 RD Sharma Solutions – Chapter 2 Exponents of Real Numbers- Exercise 2.2 | Set 1

  • Last Updated : 30 Apr, 2021

Question 1. Assuming that x, y, z are positive real numbers, simplify each of the following:

(i) \left(\sqrt{x^{-3}}\right)^5

Solution:

We have,

\left(\sqrt{x^{-3}}\right)^5

\left(\sqrt{\frac{1}{x^{3}}}\right)^5

\left(\frac{1}{x^{\frac{3}{2}}}\right)^5



\frac{1}{x^{\frac{15}{2}}}

(ii) \sqrt{x^{3}y^{-2}}

Solution:

We have,

\sqrt{x^{3}y^{-2}}

\frac{x^{\frac{3}{2}}}{y^{2×\frac{1}{2}}}

\frac{x^{\frac{3}{2}}}{y}

(iii) (x^{-\frac{2}{3}}y^{-\frac{1}{2}})^2

Solution:

We have,



(x^{-\frac{2}{3}}y^{-\frac{1}{2}})^2

\left(\frac{1}{x^{\frac{2}{3}}y^{\frac{1}{2}}}\right)^2

\frac{1}{x^{\frac{4}{3}}y}

(iv) (\sqrt{x})^{-\frac{2}{3}}\sqrt{y^4}÷\sqrt{xy^{-\frac{1}{2}}}       

Solution:

We have,

(\sqrt{x})^{-\frac{2}{3}}\sqrt{y^4}÷\sqrt{xy^{-\frac{1}{2}}}

\frac{x^{-\frac{1}{3}}y^2}{x^{\frac{1}{2}}y^{-\frac{1}{4}}}

x^{-\frac{1}{3}-\frac{1}{2}}y^{2+\frac{1}{4}}

x^{-\frac{5}{6}}y^{\frac{9}{4}}

\frac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}}



(v) \sqrt[5]{243x^{10}y^5z^{10}}

Solution:

We have,

\sqrt[5]{243x^{10}y^5z^{10}}

(243)^{\frac{1}{5}}x^{\frac{10}{5}}y^{\frac{5}{5}}z^{\frac{10}{5}}

(3^5)^{\frac{1}{5}}x^{2}yz^{2}

= 3 x2 y z2

(vi) \left(\frac{x^{-4}}{y^{-10}}\right)^\frac{5}{4}

Solution:

We have,

\left(\frac{x^{-4}}{y^{-10}}\right)^\frac{5}{4}

\frac{x^{-4×\frac{5}{4}}}{y^{-10×\frac{5}{4}}}



\frac{x^{-5}}{y^{-\frac{25}{2}}}

\frac{y^{\frac{25}{2}}}{x^5}

(vii) \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^5\left(\frac{6}{7}\right)^2

Solution:

We have,

\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^5\left(\frac{6}{7}\right)^2

\left(\frac{2^5}{3^5}\right)^\frac{1}{2}\left(\frac{6^4}{7^4}\right)^\frac{1}{2}

\left(\frac{2^5}{3^5}×\frac{6^4}{7^4}\right)^\frac{1}{2}

\left(\frac{512}{7203}\right)^\frac{1}{2}

Question 2. Simplify

(i) (16^{\frac{-1}{5}})^\frac{5}{2}       

Solution:

We have,



(16^{\frac{-1}{5}×\frac{5}{2}})

16^{\frac{-1}{2}}

(4^2)^{\frac{-1}{2}}

\frac{1}{4}

(ii) \sqrt[5]{(32)^{-3}}

Solution:

We have,

\sqrt[5]{(32)^{-3}}

(2^5)^{\frac{-3}{5}}

= 2−3

\frac{1}{8}



(iii) \sqrt[3]{(343)^{-2}}

Solution:

We have,

\sqrt[3]{(343)^{-2}}

(7^3)^{\frac{-2}{3}}

= 7−2

\frac{1}{49}

(iv) (0.001)^\frac{1}{3}

Solution:

We have,

(0.001)^\frac{1}{3}

(\frac{1}{10^3})^\frac{1}{3}



\frac{1}{10}

= 0.1

(v) \frac{25^{\frac{3}{2}}×243^\frac{3}{5}}{16^\frac{5}{4}×8^\frac{4}{3}}

Solution:

We have,

\frac{25^{\frac{3}{2}}×243^\frac{3}{5}}{16^\frac{5}{4}×8^\frac{4}{3}}

\frac{(5^2)^{\frac{3}{2}}×(3^5)^\frac{3}{5}}{(2^4)^\frac{5}{4}×(2^3)^\frac{4}{3}}

\frac{5^3×3^3}{2^5×4^2}

\frac{125×27}{32×16}

\frac{3375}{512}

(vi) \left(\frac{\sqrt{2}}{5}\right)^8÷\left(\frac{\sqrt{2}}{5}\right)^{13}       

Solution:



We have,

\left(\frac{\sqrt{2}}{5}\right)^8÷\left(\frac{\sqrt{2}}{5}\right)^{13}

\left(\frac{\sqrt{2}}{5}\right)^{8-13}

\left(\frac{\sqrt{2}}{5}\right)^{-5}

\left(\frac{5}{\sqrt{2}}\right)^5

\frac{3125}{4\sqrt{2}}

(vii) \left(\frac{5^{-1}×7^2}{5^2×7^{-4}}\right)^{\frac{7}{2}}×\left(\frac{5^{-2}×7^3}{5^3×7^{-5}}\right)^\frac{-5}{2}

Solution:

We have,

\left(\frac{5^{-1}×7^2}{5^2×7^{-4}}\right)^{\frac{7}{2}}×\left(\frac{5^{-2}×7^3}{5^3×7^{-5}}\right)^\frac{-5}{2}

\frac{5^{\frac{-7}{2}}×7^7}{5^7×7^{-14}}×\frac{5^5×7^{\frac{-15}{2}}}{5^{\frac{-15}{2}}×7^{\frac{25}{2}}}



5^{\frac{-7}{2}+5-7+\frac{15}{2}}×7^{7-\frac{15}{2}+14-\frac{25}{2}}

= 52 × 7

= 175

Question 3. Prove that

(i) (\sqrt{3×5^{-3}}÷\sqrt[3]{3^{-1}}\sqrt{5})×\sqrt[6]{3×5^6}=\frac{3}{5}

Solution:

We have,

L.H.S. = (\sqrt{3×5^{-3}}÷\sqrt[3]{3^{-1}}\sqrt{5})×\sqrt[6]{3×5^6}

((3^\frac{1}{3}×5^\frac{-3}{2})÷3^\frac{-1}{2}5^\frac{1}{2})×(3^\frac{1}{6}×5)

(3^{\frac{1}{3}+\frac{1}{2}}×5^{\frac{-3}{2}-\frac{1}{2}})×(3^\frac{1}{6}×5)

(3^{\frac{5}{6}}×5^{-2})×(3^\frac{1}{6}×5)

(3^{\frac{5}{6}+\frac{1}{6}}×5^{-2+1})



3(\frac{1}{5})

\frac{3}{5}

= R.H.S.

Hence proved.

(ii) 9^{\frac{3}{2}}-3×5^0-(\frac{1}{81})^{\frac{-1}{2}}=15

Solution:

We have,

L.H.S. = 9^{\frac{3}{2}}-3×5^0-(\frac{1}{81})^{\frac{-1}{2}}

(3^2)^{\frac{3}{2}}-3-(\frac{1}{9^2})^{\frac{-1}{2}}

(3^3)-3-(9^{-2})^{\frac{-1}{2}}

= 27 − 3 − 9

= 15

= R.H.S.

Hence proved.

(iii) \left(\frac{1}{4}\right)^{-2}-3×8^{\frac{2}{3}}×4^0+\left(\frac{9}{16}\right)^{\frac{-1}{2}}=\frac{16}{3}

Solution:

We have,

L.H.S. = \left(\frac{1}{4}\right)^{-2}-3×8^{\frac{2}{3}}×4^0+\left(\frac{9}{16}\right)^{\frac{-1}{2}}

\left(\frac{1}{2^2}\right)^{-2}-3×(2^3)^{\frac{2}{3}}+\left(\frac{3^2}{4^2}\right)^{\frac{-1}{2}}

2^4-3×4+\left(\frac{3}{4}\right)^{-1}

16-12+\frac{4}{3}

\frac{16}{3}



= R.H.S.

Hence proved.

(iv) \frac{2^{\frac{1}{2}}×3^{\frac{1}{3}}×4^{\frac{1}{4}}}{10^{\frac{-1}{5}}×5^{\frac{3}{5}}}÷\frac{3^\frac{4}{3}×5^\frac{-7}{5}}{4^\frac{-3}{5}×6}=10

Solution:

We have,

L.H.S. = \frac{2^{\frac{1}{2}}×3^{\frac{1}{3}}×4^{\frac{1}{4}}}{10^{\frac{-1}{5}}×5^{\frac{3}{5}}}÷\frac{3^\frac{4}{3}×5^\frac{-7}{5}}{4^\frac{-3}{5}×6}

\frac{2^{\frac{1}{2}}×3^{\frac{1}{3}}×(2^2)^{\frac{1}{4}}}{(2×5)^{\frac{-1}{5}}×5^{\frac{3}{5}}}÷\frac{3^\frac{4}{3}×5^\frac{-7}{5}}{(2^2)^\frac{-3}{5}×2×3}

\frac{2^{\frac{1}{2}}×3^{\frac{1}{3}}×2^{\frac{1}{2}}}{2^{\frac{-1}{5}}×5^{\frac{-1}{5}}×5^{\frac{3}{5}}}÷\frac{3^\frac{4}{3}×5^\frac{-7}{5}}{2^\frac{-6}{5}×2×3}

\frac{(2)^{\frac{1}{2}+\frac{1}{2}-\frac{6}{5}+1+\frac{1}{5}}×(3)^{\frac{1}{3}+1-\frac{4}{3}}}{(5)^{\frac{-1}{5}+\frac{3}{5}-\frac{7}{5}}}

\frac{2^1×3^0}{5^{-1}}

= 2 × 1 × 5



= 10

= R.H.S.

Hence proved.

(v) \sqrt{\frac{1}{4}}+(0.01)^{\frac{-1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}

Solution:

We have,

L.H.S. = \sqrt{\frac{1}{4}}+(0.01)^{\frac{-1}{2}}-(27)^{\frac{2}{3}}

\left(\frac{1}{2^2}\right)^\frac{1}{2}+(\frac{1}{10^2})^{\frac{-1}{2}}-(3^3)^{\frac{2}{3}}

\frac{1}{2}+10-9

\frac{3}{2}

= R.H.S.

Hence proved.

(vi) \frac{2^n+2^{n-1}}{2^{n+1}-2^n}=\frac{3}{2}

Solution:

We have,

L.H.S. = \frac{2^n+2^{n-1}}{2^{n+1}-2^n}

\frac{2^n[1+2^{-1}]}{2^n[2-1]}

= 1 + \frac{1}{2}

\frac{3}{2}

= R.H.S.

Hence proved. 

(vii) \left(\frac{64}{125}\right)^{\frac{-2}{3}}+\frac{1}{(\frac{256}{625})^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^0=\frac{61}{16}

Solution:



We have,

L.H.S. = \left(\frac{4}{5}\right)^{3×\frac{-2}{3}}+\frac{1}{(\frac{4}{5})^{4×\frac{1}{4}}}+1

\left(\frac{4}{5}\right)^{-2}+\frac{1}{(\frac{4}{5})}+1

\frac{25}{16}+\frac{5}{4}+1

\frac{61}{16}

= R.H.S.

Hence proved. 

(viii) \frac{3^{-3}×6^2×\sqrt{98}}{5^2×\sqrt[3]{\frac{1}{25}}×(15)^{\frac{-4}{3}}×3^{\frac{1}{3}}}=28\sqrt{2}

Solution:

We have,

L.H.S. = \frac{3^{-3}×6^2×\sqrt{98}}{5^2×\sqrt[3]{\frac{1}{25}}×(15)^{\frac{-4}{3}}×3^{\frac{1}{3}}}

\frac{3^{-3}×(2×3)^2×\sqrt{49×2}}{5^2×(5^2)^{\frac{-1}{3}}×(3×5)^{\frac{-4}{3}}×3^{\frac{1}{3}}}

\frac{3^{-3}×2^2×3^2×7×2^{\frac{1}{2}}}{5^2×5^{\frac{-2}{3}}×3^{\frac{-4}{3}}×5^{\frac{-4}{3}}×3^{\frac{1}{3}}}

\frac{3^{-3+2+\frac{4}{3}-\frac{1}{3}}×2^2×7×\sqrt{2}}{5^{2-\frac{2}{3}-\frac{4}{3}}}

\frac{3^0×28\sqrt{2}}{5^0}

= 28√2

= R.H.S.

Hence proved. 

(ix) \frac{(0.6)^0-(0.1)^{-1}}{(\frac{3}{8})^{-1}(\frac{3}{2})^3+(\frac{1}{3})^{-1}}=\frac{-3}{2}      

Solution:

We have,

L.H.S. = \frac{(0.6)^0-(0.1)^{-1}}{(\frac{3}{8})^{-1}(\frac{3}{2})^3+(\frac{1}{3})^{-1}}



\frac{1-10}{(\frac{8}{3})(\frac{3}{2})^3-3}

\frac{-9}{9-3}

\frac{-9}{6}

\frac{-3}{2}

= R.H.S.

Hence proved. 

Question 4. Show that

(i) \frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1

Solution:

We have,

L.H.S. = \frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}

\frac{1}{1+\frac{x^a}{x^b}}+\frac{1}{1+\frac{x^b}{x^a}}

\frac{x^b}{x^b+x^a}+\frac{x^a}{x^a+x^b}

\frac{x^a+x^b}{x^a+x^b}

= 1

= R.H.S.

Hence proved. 

(ii) \left[\left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right)÷\left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right)\right]^{a+b}=1

Solution:

We have,

L.H.S. = \left[\left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right)÷\left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right)\right]^{a+b}

\left[\left(\frac{x^{a^2-ab}}{x^{a^2+ab}}\right)÷\left(\frac{x^{b^2-ab}}{x^{b^2+ab}}\right)\right]^{a+b}

[x^{(a^2−ab)−(a^2+ab)} ÷ x^{(b^2−ab)−(b^2+ab)}]^{a+b}



= [x-2ab-(-2ab)]a+b

= [x0]a+b

= x0

= 1

= R.H.S.

Hence proved. 

(iii) \left(x^\frac{1}{a-b}\right)^{\frac{1}{a-c}}\left({x^\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left({x^\frac{1}{c-a}}\right)^{\frac{1}{c-b}}=1

Solution:

We have,

L.H.S. = \left(x^\frac{1}{a-b}\right)^{\frac{1}{a-c}}\left({x^\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left({x^\frac{1}{c-a}}\right)^{\frac{1}{c-b}}

\left(x^\frac{1}{(a-b)(a-c)}\right)\left({x^\frac{1}{(b-c)(b-a)}}\right)\left({x^\frac{1}{(c-a)(c-b)}}\right)

\left(x^{\frac{1}{(a-b)(a-c)}-\frac{1}{(b-c)(a-b)}+\frac{1}{(a-c)(b-c)}}\right)

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

= x0

= 1

= R.H.S.

Hence proved. 

(iv) \left(\frac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\frac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\frac{x^{c^2+a^2}}{x^{ac}}\right)^{a+c}=x^{2(a^3+b^3+c^3)}

Solution:

We have,

L.H.S. = \left(\frac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\frac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\frac{x^{c^2+a^2}}{x^{ac}}\right)^{a+c}

\left(x^{a^2+b^2-ab}\right)^{a+b}\left(x^{b^2+c^2-bc}\right)^{b+c}\left(x^{c^2+a^2-ac}\right)^{c+a}



\left(x^{a^3+b^3}\right)\left(x^{b^3+c^3}\right)\left(x^{a^3+c^3}\right)

x^{2(a^3+b^3+c^3)}

= R.H.S.

Hence proved. 

(v) (xa-b)a+b (xb-c)b+c (xc-a)c+a = 1

Solution:

We have,

L.H.S. = (xa-b)a+b (xb-c)b+c (xc-a)c+a 

(x^{a^2-b^2})(x^{b^2-c^2})(x^{c^2-a^2})

x^{a^2-b^2+b^2-c^2+c^2-a^2}

= x0

= 1

= R.H.S.

Hence proved. 

(vi) \left[(x^{a-a^{-1}})^{\frac{1}{a-1}}\right]^{\frac{a}{a+1}}=x

Solution:

We have,

L.H.S. = \left[(x^{a-a^{-1}})^{\frac{1}{a-1}}\right]^{\frac{a}{a+1}}

\left[x^{\frac{a(a-a^{-1})}{a^2-1}}\right]

x^{\frac{a^2-1}{a^2-1}}

= x

= R.H.S.

Hence proved. 

(vii) \left[\frac{a^{x+1}}{a^{y+1}}\right]^{x+y}\left[\frac{a^{y+2}}{a^{z+2}}\right]^{y+z}\left[\frac{a^{z+3}}{a^{z+3}}\right]^{x+z}=1

Solution:

We have,

L.H.S. = \left[\frac{a^{x+1}}{a^{y+1}}\right]^{x+y}\left[\frac{a^{y+2}}{a^{z+2}}\right]^{y+z}\left[\frac{a^{z+3}}{a^{z+3}}\right]^{x+z}

= (ax-y)x+y (ay-z)y+z (ax-z)x+z 

(a^{x^2-y^2})(a^{y^2-z^2})(a^{x^2-z^2})

a^{x^2-y^2+y^2-z^2+x^2-z^2}

= a0

= 1

= R.H.S.



Hence proved. 

(viii) \left[\frac{3^a}{3^b}\right]^{a+b}\left[\frac{3^b}{3^c}\right]^{b+c}\left[\frac{3^c}{3^a}\right]^{c+a}=1

Solution:

We have,

L.H.S. = \left[\frac{3^a}{3^b}\right]^{a+b}\left[\frac{3^b}{3^c}\right]^{b+c}\left[\frac{3^c}{3^a}\right]^{c+a}

= (3a-b)a+b (3b-c)b+c (3c-a)c+a 

(3^{a^2-b^2})(3^{b^2-c^2})(3^{c^2-a^2})

3^{a^2-b^2+b^2-c^2+c^2-a^2}

= 30

= 1

= R.H.S.

Hence proved. 

Question 5. If 2x = 3y = 12z, show that 1/z = 1/y + 2/x.

Solution:

We are given,

=> 2x = 3y = 12z = k (say)

So, we get,

=> 12 = k1/z

=> 2 × 3 × 2 = k1/z

=> 22 × 3 = k1/z

=> (k1/x)2 × (k)1/y = k1/z

=> (k)2/x × (k)1/y = k1/z

=> k^{\frac{2}{x}+\frac{1}{y}}     = k1/z 

=> 2/x + 1/y = 1/z

Hence proved.

Question 6. If 2x = 3y = 6−z, show that 1/x + 1/y + 1/z = 0.

Solution:

We are given,

=> 2x = 3y = 6−z = k (say)

So, we get,

=> 6 = k-1/z

=> (2 × 3) = k-1/z

=> k1/x × k1/y = k-1/z

=> k^{\frac{1}{x}+\frac{1}{y}}     = k-1/z

=> 1/x + 1/y = −1/z

=> 1/x + 1/y + 1/z = 0

Hence proved.

Question 7. If ax = by = cz and b2 = ac, then show that y = 2zx/(z+x).

Solution:

We are given,

=> ax = by = cz = k (say)

=> a = k1/x, b = k1/y, c = k1/z

We are given, b2 = ac

=> (k1/y)2 = k1/x × k1/z

=> k2/yk^{\frac{1}{x}+\frac{1}{z}}

=> 2/y = 1/x + 1/z

=> 2/y = (x+z)/xz

=> y = 2zx/(z+x)

Hence proved.

Question 8. If 3x = 5y = (75)z, show that z = xy/(2x+y).

Solution:

We are given,

=> 3x = 5y = (75)z = k (say)

So, we get,

=> 75 = k1/z

=> 3 × 52 = k1/z

=> (k)1/x × (k1/y)2 = k1/z

=> (k)1/x × (k)2/y = k1/z

=> k^{\frac{1}{x}+\frac{2}{y}}    = k1/z

=> 1/x + 2/y = 1/z

=> (2x+y)/xy = 1/z

=> z = xy/(2x+y)

Hence proved. 

Question 9. If (27)x = 9/3x, find x.

Solution:

We are given,

=> (27)x = 9/3x

=> (33)x = 32/3x

=> 33x = 32−x

=> 3x = 2 − x

=> 4x = 2

=> x = 2/4

=> x = 1/2

Question 10. Find the values of x in each of the following:

(i) 25x ÷ 2x = \sqrt[5]{2^{20}}

Solution:

We have,

=> 25x ÷ 2x \sqrt[5]{2^{20}}



=> 25x−x2^{20×\frac{1}{5}}

=> 24x = 24

=> 4x = 4

=> x = 1

(ii) (23)4 = (22)x

Solution:

We have,

=> (23)4 = (22)x

=> 212 = 22x

=> 2x = 12

=> x = 6

(iii) \left(\frac{3}{5}\right)^x\left(\frac{5}{3}^{2x}\right)=\frac{125}{27}

Solution:

We have,

=> \left(\frac{3}{5}\right)^x\left(\frac{5}{3}^{2x}\right)=\frac{125}{27}    

=> \frac{3^x5^{2x}}{5^{x}3^{2x}}=\frac{5^3}{3^3}

=> \frac{5^{2x-x}}{3^{2x-x}}=\frac{5^3}{3^3}

=> \frac{5^{x}}{3^{x}}=\frac{5^3}{3^3}

=> \left(\frac{5}{3}\right)^x=\left(\frac{5}{3}\right)^3

=> x = 3

(iv) 5x−2 × 32x−3 = 135

Solution:

We have,

=> 5x−2 × 32x−3 = 135

=> 5x−2 × 32x−3 = 5 × 27

=> 5x−2 × 32x−3 = 51 × 33

=> x − 2 = 1 and 2x − 3 = 3 

=> x = 3

(v) 2x−7 × 5x−4 = 1250

Solution:

We are given,

=> 2x−7 × 5x−4 = 1250

=>  2x−7 × 5x−4 = 2 × 625

=> 2x−7 × 5x−4 = 2 × 54

=> x − 7 = 1 and x − 4 = 4 

=> x = 8

(vi) (\sqrt[3]{4})^{2x+\frac{1}{2}}=\frac{1}{32}    

Solution:

We have,

=> (\sqrt[3]{4})^{2x+\frac{1}{2}}=\frac{1}{32}

=> (2^\frac{2}{3})^{2x+\frac{1}{2}}=\frac{1}{2^5}

=> (2)^{\frac{4x}{3}+\frac{1}{3}}=2^{-5}

=> 4x/3 + 1/3 = −5

=> 4x +1 = −15

=> 4x = −16

=> x = −4

(vii) 52x+3 = 1

Solution:

We have,

=> 52x+3 = 1

=> 52x+3 = 50

=> 2x + 3 = 0

=> 2x = −3

=> x = −3/2

(viii) 13^{\sqrt{x}}=4^4-3^4-6

Solution:

 We have,

=> 13^{\sqrt{x}}=4^4-3^4-6     

=> 13^{\sqrt{x}}    = 256 − 81 − 6

=> 13^{\sqrt{x}}    = 169 

=> 13^{\sqrt{x}}=13^2

=> √x = 2

=> x = 4

(ix) \left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}

Solution:

We have,

=> \left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}

=> \left(\sqrt{\frac{3}{5}}\right)^{x+1}=\left(\frac{5}{3}\right)^3



=> \left(\frac{3}{5}\right)^{\frac{x+1}{2}}=\left(\frac{3}{5}\right)^{-3}

=> (x+1)/2 = −3

=> x + 1 = −6

=> x = −7

Question 11. If x = 21/3 + 22/3, show that x3 − 6x = 6.

Solution:

Given, x = 21/3 + 22/3

Therefore, x3 = (21/3)3 + (22/3)3 + 3(21/3)(22/3)(21/3 + 22/3)

=> x3 = (21/3)3 + (22/3)3 + 3(21/3)(22/3)(x)

=> x3 = 2 + 4 + 3(2)(x)

=> x3 = 6 + 6x

=> x3 − 6x = 6

Hence proved.

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