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

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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 x²y 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) (x^a-b)^a+b (x^b-c)^b+c (x^c-a)^c+a = 1

Solution:

We have,

L.H.S. = (x^a-b)^a+b (x^b-c)^b+c (x^c-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. = $\left[\frac{a^{x+1}}{a^{y+1}}\right]^{x+y}\left[\frac{a^{y+2}}{a^{z+2}}\right]^{y+z}\left[\frac{a^{z+3}}{a^{z+3}}\right]^{x+z}$

= (a^x-y)^x+y (a^y-z)^y+z (a^x-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. = $\left[\frac{3^a}{3^b}\right]^{a+b}\left[\frac{3^b}{3^c}\right]^{b+c}\left[\frac{3^c}{3^a}\right]^{c+a}$

= (3^a-b)^a+b (3^b-c)^b+c (3^c-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 2^x = 3^y = 12^z, show that 1/z = 1/y + 2/x.

Solution:

We are given,

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

So, we get,

=> 12 = k^1/z

=> 2 × 3 × 2 = k^1/z

=> 2² × 3 = k^1/z

=> (k^1/x)² × (k)^1/y = k^1/z

=> (k)^2/x × (k)^1/y = k^1/z

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

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

Hence proved.

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

Solution:

We are given,

=> 2^x = 3^y = 6^−z = k (say)

So, we get,

=> 6 = k^-1/z

=> (2 × 3) = k^-1/z

=> k^1/x × k^1/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 a^x = b^y = c^z and b² = ac, then show that y = 2zx/(z+x).

Solution:

We are given,

=> a^x = b^y = c^z = k (say)

=> a = k^1/x, b = k^1/y, c = k^1/z

We are given, b² = ac

=> (k^1/y)² = k^1/x × k^1/z

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

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

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

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

Hence proved.

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

Solution:

We are given,

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

So, we get,

=> 75 = k^1/z

=> 3 × 5² = k^1/z

=> (k)^1/x × (k^1/y)² = k^1/z

=> (k)^1/x× (k)^2/y = k^1/z

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

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

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

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

Hence proved.

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

Solution:

We are given,

=> (27)^x = 9/3^x

=> (3³)^x = 3²/3^x

=> 3^3x = 3^2−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) 2^5x ÷ 2^x = $\sqrt[5]{2^{20}}$

Solution:

We have,

=> 2^5x ÷ 2^x= $\sqrt[5]{2^{20}}$

=> 2^5x−x = $2^{20×\frac{1}{5}}$

=> 2^4x = 2⁴

=> 4x = 4

=> x = 1

(ii) (2³)⁴= (2²)^x

Solution:

We have,

=> (2³)⁴ = (2²)^x

=> 2¹² = 2^2x

=> 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) 5^x−2× 3^2x−3 = 135

Solution:

We have,

=> 5^x−2 × 3^2x−3 = 135

=> 5^x−2 × 3^2x−3 = 5 × 27

=> 5^x−2× 3^2x−3 = 5¹ × 3³

=> x − 2 = 1 and 2x − 3 = 3

=> x = 3

(v) 2^x−7× 5^x−4 = 1250

Solution:

We are given,

=> 2^x−7 × 5^x−4 = 1250

=> 2^x−7× 5^x−4= 2 × 625

=> 2^x−7 × 5^x−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) 5^2x+3 = 1

Solution:

We have,

=> 5^2x+3 = 1

=> 5^2x+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 = 2^1/3 + 2^2/3, show that x³ − 6x = 6.

Solution:

Given, x = 2^1/3 + 2^2/3

Therefore, x³ = (2^1/3)³ + (2^2/3)³+ 3(2^1/3)(2^2/3)(2^1/3 + 2^2/3)

=> x³= (2^1/3)3 + (2^2/3)³ + 3(2^1/3)(2^2/3)(x)

=> x³ = 2 + 4 + 3(2)(x)

=> x³ = 6 + 6x

=> x³ − 6x = 6

Hence proved.

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Last Updated : 30 Apr, 2021

Class 9 RD Sharma Solutions - Chapter 2 Exponents of Real Numbers- Exercise 2.1

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

Chapter 1: Number Systems

Chapter 2: Exponents and Powers of Real Numbers

Chapter 3: Rationalisation

Chapter 4: Algebraic Identities

Chapter 5: Factorization of Algebraic Expressions

Chapter 6: Factorization of Polynomials

Chapter 7: Introduction to Euclidâ€™s Geometry

Chapter 8: Lines and Angles

Chapter 9: Triangle and Its Angles

Chapter 10: Congruent Triangles

Chapter 11: Coordinate Geometry

Chapter 12: Heronâ€™s Formula

Chapter 13: Linear Equations in Two Variables

Chapter 14: Quadrilaterals

Chapter 15: Areas of Parallelograms and Triangles

Chapter 16: Circles

Chapter 17: Constructions

Chapter 18: Surface Area and Volume of a Cuboid and a Cube

Chapter 19: Surface Area and Volume of a Right Circular Cylinder

Chapter 20: Surface Area and Volume of a Right Circular Cone

Chapter 21: Surface Area and Volume of a Sphere

Chapter 22: Tabular Representation of Statistical Data

Chapter 23: Graphical Representation of Statistical Data

Chapter 24: Measures of Central Tendency

Chapter 25: Probability

Chapter 1: Number Systems

Chapter 2: Exponents and Powers of Real Numbers

Chapter 3: Rationalisation

Chapter 4: Algebraic Identities

Chapter 5: Factorization of Algebraic Expressions

Chapter 6: Factorization of Polynomials

Chapter 7: Introduction to Euclidâ€™s Geometry

Chapter 8: Lines and Angles

Chapter 9: Triangle and Its Angles

Chapter 10: Congruent Triangles

Chapter 11: Coordinate Geometry

Chapter 12: Heronâ€™s Formula

Chapter 13: Linear Equations in Two Variables

Chapter 14: Quadrilaterals

Chapter 15: Areas of Parallelograms and Triangles

Chapter 16: Circles

Chapter 17: Constructions

Chapter 18: Surface Area and Volume of a Cuboid and a Cube

Chapter 19: Surface Area and Volume of a Right Circular Cylinder

Chapter 20: Surface Area and Volume of a Right Circular Cone

Chapter 21: Surface Area and Volume of a Sphere

Chapter 22: Tabular Representation of Statistical Data

Chapter 23: Graphical Representation of Statistical Data

Chapter 24: Measures of Central Tendency

Chapter 25: Probability

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

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Question 2. Simplify

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Question 3. Prove that

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Question 4. Show that

(i)

(ii)

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

(ii) $\sqrt{x^{3}y^{-2}}$

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

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

(v) $\sqrt[5]{243x^{10}y^5z^{10}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(v) (x^a-b)^a+b (x^b-c)^b+c (x^c-a)^c+a = 1

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

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

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

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

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

Question 7. If a^x = b^y = c^z and b² = ac, then show that y = 2zx/(z+x).

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

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

(i) 2^5x ÷ 2^x = $\sqrt[5]{2^{20}}$

(ii) (2³)⁴= (2²)^x

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

(iv) 5^x−2× 3^2x−3 = 135

(v) 2^x−7× 5^x−4 = 1250

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

(vii) 5^2x+3 = 1

(viii) $13^{\sqrt{x}}=4^4-3^4-6$

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

Question 11. If x = 2^1/3 + 2^2/3, show that x³ − 6x = 6.