OP Malhotra Class 9 Maths Solutions Chapter 6 Indices/Exponents Ex 6(C)

Effective OP Malhotra Class 9 Maths Solutions Chapter 6 Indices/Exponents Ex 6(C) can help bridge the gap between theory and application.

S Chand Class 9 ICSE Maths Solutions Chapter 6 Indices/Exponents Ex 6(C)

Question 1.
Evaluate:
(i) 5° x 4^-1 + 8^-1
(ii) \(\sqrt[3]{(64)^{-4}(125)^{-2}}\)
(iii) \(\left(9^{-3} \times 16^{3.2}\right)^{1,6}\)
(iv) \(\sqrt[3]{(16)^{-34} \times(125)^{-2}}\)
(v) \((32)^{-2 / 5} \quad(216)^{-23}\)
Solution:
(i) \(5^0 \times 4^{-1}+8^{1 / 3}=1 \times \frac{1}{4}+(2 \times 2 \times 2)^{1 / 3}\)
= \(\frac{1}{4}+\left(2^3\right)^{1 / 3}=\frac{1}{4}+2^{3 \times 1 / 3}\)
= \(\frac{1}{4}+2^1=2+\frac{1}{4}=2 \frac{1}{4}\)

(ii) \(\sqrt[3]{(64)^{-4}(125)^{-2}}\)
= \(\left[(64)^{-4}(125)^{-2}\right]^{1 / 3}=(64)^{-4 \times 1 / 3} \cdot(125)^{-2 \times 1 / 3}\)
= \((64)^{-4 / 3} \cdot(125)^{-2 / 3}=\left(4^3\right)^{-4 / 3} \times\left(5^3\right)^{-2 / 3}\)
= \(4^{3 \times(-4) 3)} \times 5^{3 \times(-2 / 3)}=4^{-4} \times 5^{-2}\)
= \(\frac{1}{4^4 \times 5^2}=\frac{1}{4 \times 4 \times 4 \times 4 \times 5 \times 5}=\frac{1}{6400}\)

(iii) \(\left(9^{-3} \times 16^{3 / 2}\right)^{1 / 6}=\left[\left(3^2\right)^{-3} \times\left(2^4\right)^{3 / 2}\right]^{1 / 6}\)
= \(\left[(3)^{-6} \times(2)^6\right]^{16}=3^{-6.1 .6} \times 2^{6 \times 1 / 6}\)
= \(3^1 \times 2^{\prime}=\frac{1}{3} \times 2=\frac{2}{3}\)

(iv) \(\sqrt[3]{(16)^{-3 / 4} \times(125)^{-2}}=\sqrt[3]{\left(2^4\right)^{-3 / 4} \times\left(5^3\right)^{-2}}\)
= \(\left[2^{4 \times(-3.4)} \times 5^{3 \times(-2)}\right]^{13}\)
= \(\left[2^{-3} \times 5^{-6}\right]^{1 / 3}=\left[\frac{1}{2^3 \times 5^6}\right]^{\frac{1}{3}}\)
= \(\frac{1}{\left(2^3\right)^{1 / 3} \times\left(5^6\right)^{\frac{1}{3}}}=\frac{1}{2^1 \times 5^2}\)
= \(\frac{1}{2 \times 5 \times 5}=\frac{1}{50}\)

(v) (32)^-2/5 + (216)^-2/3
= (2⁵)^-2/5 + (6³)^-2/3
= \((2)^{-2 / 5 \times 5}+(6)^{3 \times(-2 / 3)}\)
= 2^-2 + 6^-2
= \(\left(\frac{2}{6}\right)^{-2}=\left(\frac{6}{2}\right)^2=\frac{6 \times 6}{2 \times 2}=\frac{36}{4}\) = 9

Question 2.
(i) \(\left(\frac{64}{125}\right)^{-2 / 3}+\frac{1}{\left(\frac{256}{625}\right)^{1 / 4}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^0\)
(ii) \(9^{3 / 2}-3 \times(5)^0-\left(\frac{1}{81}\right)^{-1 / 2}\)
(iii) \(\left(\frac{1}{4}\right)^{-2}-3(8)^{2 / 3} \times 4^0+\left(\frac{9}{16}\right)^{-1 / 2}\)
(iv) \(16^{3,4}+2\left(\frac{1}{2}\right)^{-1}\)
(v) \((81)^{3 / 4}-\left(\frac{1}{32}\right)^{-2 / 5}+(8)^{1 / 3}\left(\frac{1}{2}\right)^{-2}\)
Solution:

Question 3.
(a) Evaluate : x^1/2 x y^-1 x z^2/3 when x = 9, y = 2, and z = 8.
(b) Evaluate as a fraction :
\(\left(\frac{27}{8}\right)^{2.3}-\left(\frac{1}{4}\right)^{-2}+(5)^0\)
(c) Evaluate as a fraction :
\(\sqrt{\frac{1}{4}}+(0.01)^{-1.2}-(27)^{2.3}\)
Solution:
(a) \(x^{1.2} \times y^{-1} \times z^{2 / 3}\), x = 9, y = 2, z = 8
= (9)\(^{1 / 2} \times(2)^{-1} \times(8)^{2.3}=\left(3^2\right)^{12} \times \frac{1}{2} \times\left(2^3\right)^{2,3}\)
= \(3^{2,12} \times \frac{1}{2} \times 2^{3 \cdot 2 \cdot 3}\)
= \(3^1 \times \frac{1}{2} \times 2^2=3 \times \frac{1}{2} \times 4\)
= 3 x 2 = 6

Question 4.
Simplify:
(i) \(\frac{(64)^{\frac{5 n}{6}} \cdot(27)^{-\frac{n}{6}}}{(12)^{-n \cdot 2}} \cdot 2^{6 n}\)
(ii) \(\frac{6 .(8)^{n+1}+16.2^{3 n-2}}{10.2^{3 n+1}-7 .(8)^n}\)
(iii) \(\frac{4^{2 n} \cdot 2^{n+1}}{2^{n-3} \cdot 4^{2 n+1}}\)
(iv) \(\frac{(5)^{2 n+3}-(25)^{n+2}}{\left[(125)^{n+1}\right\}^{1 / 3}}\)
Solution:

Question 5.
If \(\frac{49^{n+1} \cdot 7^n-(343)^n}{7^{3 m} \cdot 2^n}=\frac{3}{343}\), prove that m = (m + 1)
Solution:

Question 6.
If 2205 = 3^a x 5^b x 7^c
(i) Find the numerical values of a, b and c
(ii) Hence evaluate 3^a x 5^-b x 7^-c
Solution:
(i) 2205 = 3^a x 5^b x 7^c
3².5¹.7² = 3^a.5^b.7^c

Comparing we get,
a = 2, b = 1, c = 2

(ii) Now 3^a x 5^-b x 7^-c = \(\frac{3^a}{5^b \times 7^c}=\frac{3^2}{5^1 \times 7^2}\)
= \(\frac{9}{5 \times 49}=\frac{9}{245}\)

Question 7.
If \(\left[\frac{b^3 c^{-2}}{b^{-4} c^3}\right]^{-3}+\left[\frac{b^{-1} c}{b^2 \cdot c^{-2}}\right]^5\) = b^x.c^x, prove that x + y + 6 = 0
Solution:
\(\left[\frac{b^3 c^{-2}}{b^{-4} c^3}\right]^{-3}+\left[\frac{b^{-1} c}{b^2 \cdot c^{-2}}\right]^5=b^4 \cdot c^1\)
⇒ \(\frac{b^{-9} \cdot c^0}{b^{12} \cdot c^{-9}} \div \frac{b^{-5} c^5}{b^{10} \cdot c^{-10}}=b^x \cdot c^1\)
⇒ \(\frac{b^{-9} \cdot c^6}{b^{12} \cdot c^{-9}} \times \frac{b^{10} c^{-10}}{b^{-5} c^5}=b^{\mathrm{r}} \cdot c^1\)
⇒ \(b^{-9-12+10+5} \cdot c^{6+9-10-15}=b^4 \cdot c^5\)
⇒ \(b^{-21+15} \cdot c^{15-15}=b^x \cdot c^1\)
⇒ \(b^{-6} \cdot c^0=b^\tau \cdot c^v\)
Comparing both sides, x = – 6, y = 0
Now R.H.S. = x + y + 6 = – 6 + 6 = 0
= R.H.S.

Question 8.
prove that
(i) \((x+y)^{-1}\left(x^{-1}+y^{-1}\right)=\frac{1}{x y}\)
(ii) \(\left(x^{-1}+y^1\right)^{-1}=\frac{x y}{x+y}\)
(iii) \(\frac{a+b+c}{a^{-1} b^{-1}+b^{-1} c^{-1}+c^{-1} a^{-1}}=a b c \text {. }\)
(iv) \(\begin{aligned}
\frac{1}{1+x^{b-a}+x^{c-a}}+ & \frac{1}{1+x^{a-b}+x^{c-b}} \\
& +\frac{1}{1+x^{b-c}+x^{a-c}}=1 .
\end{aligned}\)
(v) \(\sqrt[a b]{\frac{x^a}{x^b}} \cdot \sqrt[b]{\frac{x^b}{x^c}} \cdot \sqrt[c a]{\frac{x^c}{x^a}}=1\)
Solution:

Question 9.
Simplify
(i) \(\left(p^{\frac{1}{3}}-p^{\frac{-1}{3}}\right)\left(p^{\frac{2}{3}}+1+p^{\frac{-2}{3}}\right)\)
(ii) \((\sqrt{11}+\sqrt{3})^{\frac{1}{3}}(\sqrt{11}-\sqrt{3})^{\frac{1}{3}}\)
Solution:

Question 10.
If 3^x = 5^y = 45^z, prove that x = \(\frac{2 y z}{y-z}\)
Solution:
3^x = 5^y = 45^z = a (Suppose)
∴ a^{\(\frac { 1 }{ x }\)} = 3, a^{\(\frac { 1 }{ y }\)}, a^{\(\frac { 1 }{ z }\)} = 45
45 = 3 x 3 x 5
⇒ 45 = 3² x 5
⇒ \(a^{\frac{1}{z}}=a^{\frac{2}{x} \times} a^{\frac{1}{y}}\)
⇒ \(a^{\frac{1}{z}}=a^{\frac{2}{x}+\frac{1}{y}}\)
Comparing we get,
\(\frac{1}{z}=\frac{2}{x}+\frac{1}{y} \Rightarrow \frac{2}{x}=\frac{1}{z}-\frac{1}{y}\)
⇒ \(\frac{2}{x}=\frac{y-z}{y z} \Rightarrow \frac{1}{x}=\frac{y-z}{2 y z}\)
∴ x = \(\frac { 2yz }{ y-z }\)
Hence proved.

Question 11.
If x = ab^p-1, y = ab^q-1 and z = ab^r-1 prove that x^p-r x y^r-p.z^p-q = 1.
Solution:
x = ab^p-1, y = ab^q-1 and z = ab^r-1
L.H.S. = x^p-r x y^r-p.z^p-q = 1
= (ab^p-1)^p-r.(ab^q-1)^r-p.(ab^r-1)^p-q
= a^p-r.b^(q-r)(p-1).a^r-p.b^(r-p)(q-1).a^p-q.b^(p-q)(r-1)
= a^q-r+r-p+p-r.b^pq-q-rp+r.b^rq-r-pq+p.b^pr-p-qr+q
= a°.b^{pq-q-r p+r+r q-r-pq+p+pr-p-qr+q}
= a°.b° = 1 x 1 = 1 = R.H.S.