RD Sharma 2020 solution class 9 chapter 2 Exponents of Real Numbers Exercise 2.1

Exercise 2.1

Page-2.12

Question 1:

Simplify the following:

(i) $3{\left(a^4{b}^3\right)}^{10}\times 5{\left(a^2{b}^2\right)}^3$

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

(iii) $\frac{\left(4\times {10}^7\right)\left(6\times {10}^{-5}\right)}{8\times {10}^4}$

(iv) $\frac{4a{b}^2\left(-5a{b}^3\right)}{10a^2{b}^2}$

(v) ${\left(\frac{x^2{y}^2}{a^2{b}^3}\right)}^n$

(vi) $\frac{{\left(a^{3n-9}\right)}^6}{a^{2n-4}}$

Answer 1:

(i)
$$\begin{gathered} 3{\left(a^4{b}^3\right)}^{10}\times 5{\left(a^2{b}^2\right)}^3 \\ =3\times a^{40}\times b^{30}\times 5\times a^6\times b^6 \\ =15\times a^{40}\times a^6\times b^{30}\times b^6 \\ =15\times a^{40+6}\times b^{30+6} \left[a^m\times a^n=a^{m+n}\right] \\ =15a^{46}{b}^{36} \end{gathered}$$

(ii)
$$\begin{gathered} {\left(2x^{-2}{y}^3\right)}^3 \\ =2^3\times {\left(x^{-2}\right)}^3\times {\left(y^3\right)}^3 \\ =8\times x^{-6}\times y^9 \\ =8x^{-6}{y}^9 \end{gathered}$$

(iii)
$$\begin{gathered} \frac{\left(4\times {10}^7\right)\left(6\times {10}^{-5}\right)}{8\times {10}^4} \\ =\frac{4\times {10}^7\times 6\times {10}^{-5}}{8\times {10}^4} \\ =\frac{24\times {10}^{7+\left(-5\right)}}{8\times {10}^4} \\ =\frac{24\times {10}^2}{8\times {10}^4} \end{gathered}$$
$$\begin{gathered} =\frac{24\times {10}^2\times {10}^{-4}}{8} \\ =3\times {10}^{2+\left(-4\right)} \\ =3\times {10}^{-2} \\ =\frac{3}{100} \end{gathered}$$

(iv)
$$\begin{gathered} \frac{4a{b}^2\left(-5a{b}^3\right)}{10a^2{b}^2} \\ =\frac{4\times a\times b^2\times \left(-5\right)\times a\times b^3}{10a^2{b}^2} \\ =\frac{-20\times a^1\times a^1\times b^2\times b^3}{10a^2{b}^2} \end{gathered}$$
$$\begin{gathered} =\frac{-20\times a^{1+1}\times b^{2+3}}{10a^2{b}^2} \\ =-2\times a^2\times b^5\times a^{-2}\times b^{-2} \\ =-2\times a^{2+\left(-2\right)}\times b^{5+\left(-2\right)} \\ =-2\times a^0\times b^3 \\ =-2b^3 \end{gathered}$$

(v)
$$\begin{gathered} {\left(\frac{x^2{y}^2}{a^2{b}^3}\right)}^n \\ =\frac{{\left(x^2\right)}^n{\left(y^2\right)}^n}{{\left(a^2\right)}^n{\left(b^3\right)}^n} \\ =\frac{x^{2n}{y}^{2n}}{a^{2n}{b}^{3n}} \end{gathered}$$

(vi)
$$\begin{gathered} \frac{{\left(a^{3n-9}\right)}^6}{a^{2n-4}} \\ =\frac{a^{6\left(3n-9\right)}}{a^{2n-4}} \\ =\frac{a^{\left(18n-54\right)}}{a^{2n-4}} \end{gathered}$$
$$\begin{gathered} =a^{\left(18n-54\right)}\times a^{-\left(2n-4\right)} \\ =a^{18n-54}\times a^{-2n+4} \\ =a^{18n-54-2n+4} \\ =a^{16n-50} \end{gathered}$$

Question 2:

If $a=3$ and $b=-2$, find the values of:
(i) $a^a+b^b$
(ii) $a^b+b^a$
(iii) ${\left(a+b\right)}^{ab}$

Answer 2:

(i) $a^a+b^b$
Here, $a=3$ and $b=-2$.
Put the values in the expression $a^a+b^b$.
$$\begin{gathered} 3^3+{\left(-2\right)}^{-2} \\ =27+\frac{1}{{\left(-2\right)}^2} \\ =27+\frac{1}{4} \\ =\frac{108+1}{4} \\ =\frac{109}{4} \end{gathered}$$

(ii) $a^b+b^a$
Here, $a=3$ and $b=-2$.
Put the values in the expression $a^b+b^a$.
$$\begin{gathered} 3^{-2}+{\left(-2\right)}^3 \\ ={\left(\frac{1}{3}\right)}^2+\left(-8\right) \\ =\frac{1}{9}-8 \\ =\frac{1-72}{9} \\ =-\frac{71}{9} \end{gathered}$$

(iii) ${\left(a+b\right)}^{ab}$
Here, $a=3$ and $b=-2$.
Put the values in the expression ${\left(a+b\right)}^{ab}$.
$$\begin{gathered} {\left[3+\left(-2\right)\right]}^{3\left(-2\right)} \\ ={\left(1\right)}^{-6} \\ =1 \end{gathered}$$

Question 3:

Prove that:

(i) ${\left(\frac{x^a}{x^b}\right)}^{a^2+ab+b^2}\times {\left(\frac{x^b}{x^c}\right)}^{b^2+bc+c^2}\times {\left(\frac{x^c}{x^a}\right)}^{c^2+ca+a^2}=1$
(ii) ${\left(\frac{x^a}{x^b}\right)}^c\times {\left(\frac{x^b}{x^c}\right)}^a\times {\left(\frac{x^c}{x^a}\right)}^b=1$

Answer 3:

(i) ${\left(\frac{x^a}{x^b}\right)}^{a^2+ab+b^2}\times {\left(\frac{x^b}{x^c}\right)}^{b^2+bc+c^2}\times {\left(\frac{x^c}{x^a}\right)}^{c^2+ca+a^2}=1$

Consider the left hand side:
$$\begin{gathered} {\left(\frac{x^a}{x^b}\right)}^{a^2+ab+b^2}\times {\left(\frac{x^b}{x^c}\right)}^{b^2+bc+c^2}\times {\left(\frac{x^c}{x^a}\right)}^{c^2+ca+a^2} \\ =\frac{x^{a\left(a^2+ab+b^2\right)}}{x^{b\left(a^2+ab+b^2\right)}}\times \frac{x^{b\left(b^2+bc+c^2\right)}}{x^{c\left(b^2+bc+c^2\right)}}\times \frac{x^{c\left(c^2+ca+a^2\right)}}{x^{a\left(c^2+ca+a^2\right)}} \\ =x^{a\left(a^2+ab+b^2\right)-b\left(a^2+ab+b^2\right)}\times x^{b\left(b^2+bc+c^2\right)-c\left(b^2+bc+c^2\right)}\times x^{c\left(c^2+ca+a^2\right)-a\left(c^2+ca+a^2\right)} \\ =x^{\left(a-b\right)\left(a^2+ab+b^2\right)}\times x^{\left(b-c\right)\left(b^2+bc+c^2\right)}\times x^{\left(c-a\right)\left(c^2+ca+a^2\right)} \\ =x^{\left(a^3-b^3\right)}\times x^{\left(b^3-c^3\right)}\times x^{\left(c^3-a^3\right)} \\ =x^{\left(a^3-b^3+b^3-c^3+c^3-a^3\right)} \\ =x^0 \\ =1 \end{gathered}$$
Left hand side is equal to right hand side.
Hence proved.

(ii) ${\left(\frac{x^a}{x^b}\right)}^c\times {\left(\frac{x^b}{x^c}\right)}^a\times {\left(\frac{x^c}{x^a}\right)}^b=1$
Consider the left hand side:
$$\begin{gathered} =\frac{x^{ac}}{x^{bc}}\times \frac{x^{ba}}{x^{ca}}\times \frac{x^{cb}}{x^{ab}} \\ =\frac{x^{ac}}{x^{bc}}\times \frac{x^{ba}}{x^{ca}}\times \frac{x^{cb}}{x^{ab}} \\ =\frac{x^{ac}\times x^{ba}\times x^{cb}}{x^{bc}\times x^{ca}\times x^{ab}} \\ =\frac{x^{ac+ba+cb}}{x^{bc+ca+ab}} \\ =1 \end{gathered}$$
Left hand side is equal to right hand side.
Hence proved.

Question 4:

Prove that:

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

(ii) $\frac{{\displaystyle 1}}{{\displaystyle 1+x^{b-a}+x^{c-a}}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1$

Answer 4:

(i) Consider the left hand side:
$$\begin{gathered} \frac{1}{1+x^{a-b}}+\frac{{\displaystyle 1}}{{\displaystyle 1+x^{b-a}}} \\ =\frac{1}{1+{\displaystyle \frac{x^a}{x^b}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+\frac{{\displaystyle x^b}}{x^a}}} \\ =\frac{1}{{\displaystyle \frac{x^b+x^a}{x^b}}}+\frac{{\displaystyle 1}}{{\displaystyle \frac{{\displaystyle x^a+x^b}}{x^a}}} \\ =\frac{x^b}{{\displaystyle x^b+x^a}}+\frac{{\displaystyle x^a}}{{\displaystyle x^a+x^b}} \\ =\frac{x^b+x^a}{{\displaystyle x^b+x^a}} \\ =1 \end{gathered}$$
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
$$\begin{gathered} \frac{{\displaystyle 1}}{{\displaystyle 1+x^{b-a}+x^{c-a}}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}} \\ =\frac{{\displaystyle 1}}{{\displaystyle 1+\frac{{\displaystyle x^b}}{x^a}+\frac{{\displaystyle x^c}}{x^a}}}+\frac{1}{1+{\displaystyle \frac{x^a}{x^b}}+{\displaystyle \frac{x^c}{x^b}}}+\frac{1}{1+{\displaystyle \frac{x^b}{x^c}}+{\displaystyle \frac{x^a}{x^c}}} \\ =\frac{{\displaystyle 1}}{{\displaystyle \frac{{\displaystyle x^a+x^b+x^c}}{x^a}}}+\frac{1}{{\displaystyle \frac{x^b+x^a+x^c}{x^b}}}+\frac{1}{{\displaystyle \frac{x^c+x^b+x^c}{x^c}}} \\ =\frac{{\displaystyle x^a}}{{\displaystyle x^a+x^b+x^c}}+\frac{x^b}{{\displaystyle x^b+x^a+x^c}}+\frac{x^c}{{\displaystyle x^c+x^b+x^c}} \\ =\frac{x^a+x^b+x^c}{{\displaystyle x^a+x^b+x^c}} \\ =1 \end{gathered}$$
Therefore left hand side is equal to the right hand side. Hence proved.

Question 5:

Prove that:

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

(ii) ${\left(a^{-1}+b^{-1}\right)}^{-1}=\frac{ab}{a+b}$

Answer 5:

(i) Consider the left hand side:
$$\begin{gathered} \frac{a+b+c}{a^{-1}{b}^{-1}+b^{-1}{c}^{-1}+c^{-1}{a}^{-1}} \\ =\frac{a+b+c}{{\displaystyle \frac{1}{ab}}+{\displaystyle \frac{1}{bc}}+{\displaystyle \frac{1}{ca}}} \\ =\frac{a+b+c}{{\displaystyle \frac{c+a+b}{abc}}} \\ =\left(a+b+c\right)\times \left(\frac{abc}{a+b+c}\right) \\ =abc \end{gathered}$$
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
$$\begin{gathered} {\left(a^{-1}+b^{-1}\right)}^{-1} \\ =\frac{1}{a^{-1}+b^{-1}} \\ =\frac{1}{{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}} \\ =\frac{1}{{\displaystyle \frac{b+a}{ab}}} \\ =\frac{ab}{a+b} \end{gathered}$$
Therefore left hand side is equal to the right hand side. Hence proved.

Question 6:

If abc = 1, show that $\frac{1}{1+a+b^{-1}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+c^{-1}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+c+a^{-1}}}=1$.

Answer 6:

Consider the left hand side:
$$\begin{gathered} \frac{1}{1+a+b^{-1}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+c^{-1}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+c+a^{-1}}} \\ =\frac{1}{1+a+{\displaystyle \frac{1}{b}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+\frac{1}{c}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+c+\frac{1}{a}}} \\ =\frac{1}{{\displaystyle \frac{b+ab+1}{b}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+ab}}+\frac{{\displaystyle 1}}{{\displaystyle 1+\frac{{\displaystyle 1}}{ab}+\frac{1}{a}}} \left(abc=1\right) \\ =\frac{b}{{\displaystyle b+ab+1}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+ab}}+\frac{{\displaystyle ab}}{{\displaystyle ab+1+b}} \\ =\frac{b+1+ab}{{\displaystyle b+ab+1}} \\ =1 \end{gathered}$$
Hence proved.

Question 7:

Simplify the following:

(i) $\frac{3^n\times 9^{n+1}}{3^{n-1}\times 9^{n-1}}$

(ii) $\frac{5\times {25}^{n+1}-25\times 5^{2n}}{5\times 5^{2n+3}-{25}^{n+1}}$

(iii) $\frac{5^{n+3}-6\times 5^{n+1}}{9\times 5^x-2^2\times 5^n}$

(iv) $\frac{6{\left(8\right)}^{n+1}+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7{\left(8\right)}^n}$

Answer 7:

(i)
$$\begin{gathered} \frac{3^n\times 9^{n+1}}{3^{n-1}\times 9^{n-1}} \\ =\frac{3^n\times {\left(3^2\right)}^{\left(n+1\right)}}{3^{n-1}\times {\left(3^2\right)}^{n-1}} \\ =\frac{3^n\times 3^{2n+2}}{3^{n-1}\times 3^{2n-2}} \end{gathered}$$
$$\begin{gathered} =\frac{3^{n+2n+2}}{3^{n-1+2n-2}} \\ =\frac{3^{3n+2}}{3^{3n-3}} \\ =3^{3n+2-3n+3} \\ =3^5 \\ =243 \end{gathered}$$

(ii)
$$\begin{gathered} \frac{5\times {25}^{n+1}-25\times 5^{2n}}{5\times 5^{2n+3}-{25}^{n+1}} \\ =\frac{5\times {\left(5^2\right)}^{n+1}-\left(5^2\right)\times 5^{2n}}{5\times 5^{2n+3}-{\left(5^2\right)}^{n+1}} \\ =\frac{5\times \left(5^{2n+2}\right)-\left(5^2\right)\times 5^{2n}}{5\times 5^{2n+3}-\left(5^{2n+2}\right)} \\ =\frac{5^{1+2n+2}-5^{2+2n}}{5^{1+2n+3}-5^{2n+2}} \end{gathered}$$
$$\begin{gathered} =\frac{5^{2n+3}-5^{2+2n}}{5^{2n+4}-5^{2n+2}} \\ =\frac{5^{2+2n}\left(5-1\right)}{5^{2n+2}\left(5^2-1\right)} \\ =\frac{4}{24} \\ =\frac{1}{6} \end{gathered}$$

(iii)
$$\begin{gathered} \frac{5^{n+3}-6\times 5^{n+1}}{9\times 5^n-2^2\times 5^n} \\ =\frac{5^{n+1}\left(5^2-6\right)}{5^n\left(9-2^2\right)} \\ =\frac{5^n\times 5\times \left(25-6\right)}{5^n\left(9-4\right)} \\ =\frac{5\times 19}{5} \\ =19 \end{gathered}$$

(iv)
$$\begin{gathered} \frac{6{\left(8\right)}^{n+1}+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7{\left(8\right)}^n} \\ =\frac{6{\left(2^3\right)}^{n+1}+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7{\left(2^3\right)}^n} \\ =\frac{6\left(2^{3n+3}\right)+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7\left(2^{3n}\right)} \end{gathered}$$
$$\begin{gathered} =\frac{6\times 2^{3n}\left(2^3\right)+16{\left(2\right)}^{3n}{2}^{-2}}{10{\left(2\right)}^{3n}\left(2^1\right)-7\left(2^{3n}\right)} \\ =\frac{2^{3n}\left(48+4\right)}{2^{3n}\left(20-7\right)} \\ =\frac{52}{13} \\ =4 \end{gathered}$$

Question 8:

Solve the following equations for x:

(i) $7^{2x+3}=1$

(ii) $2^{x+1}=4^{x-3}$

(iii) $2^{5x+3}=8^{x+3}$

(iv) $4^{2x}=\frac{1}{32}$

(v) $4^{x-1}\times {\left(0.5\right)}^{3-2x}={\left(\frac{1}{8}\right)}^x$

(vi) $2^{3x-7}=256$

Answer 8:

(i)
$$\begin{gathered} 7^{2x+3}=1 \\ \Rightarrow 7^{2x+3}=7^0 \\ \Rightarrow 2x+3=0 \\ \Rightarrow 2x=-3 \\ \Rightarrow x=-\frac{3}{2} \end{gathered}$$

(ii)
$$\begin{gathered} 2^{x+1}=4^{x-3} \\ \Rightarrow 2^{x+1}={\left(2^2\right)}^{x-3} \\ \Rightarrow 2^{x+1}=\left(2^{2x-6}\right) \\ \Rightarrow x+1=2x-6 \\ \Rightarrow x=7 \end{gathered}$$

(iii)
$$\begin{gathered} 2^{5x+3}=8^{x+3} \\ \Rightarrow 2^{5x+3}={\left(2^3\right)}^{x+3} \\ \Rightarrow 2^{5x+3}=2^{3x+9} \\ \Rightarrow 5x+3=3x+9 \\ \Rightarrow 2x=6 \\ \Rightarrow x=3 \end{gathered}$$

(iv)
$$\begin{gathered} 4^{2x}=\frac{1}{32} \\ \Rightarrow {\left(2^2\right)}^{2x}=\frac{1}{2^5} \\ \Rightarrow 2^{4x}\times 2^5=1 \\ \Rightarrow 2^{4x+5}=2^0 \\ \Rightarrow 4x+5=0 \\ \Rightarrow x=-\frac{5}{4} \end{gathered}$$

(v)
$$\begin{gathered} 4^{x-1}\times {\left(0.5\right)}^{3-2x}={\left(\frac{1}{8}\right)}^x \\ \Rightarrow {\left(2^2\right)}^{x-1}\times {\left(\frac{1}{2}\right)}^{3-2x}={\left(\frac{1}{2^3}\right)}^x \\ \Rightarrow {\left(2^2\right)}^{x-1}\times {\left(2^{-1}\right)}^{3-2x}={\left(2^{-3}\right)}^x \\ \Rightarrow 2^{2x-2}\times 2^{2x-3}=2^{-3x} \\ \Rightarrow 2^{2x-2+2x-3}=2^{-3x} \\ \Rightarrow 2^{4x-5}=2^{-3x} \\ \Rightarrow 4x-5= -3x \\ \Rightarrow 7x= 5 \\ \Rightarrow x= \frac{5}{7} \end{gathered}$$

(vi)
$$\begin{gathered} 2^{3x-7}=256 \\ \Rightarrow 2^{3x-7}=2^8 \\ \Rightarrow 3x-7=8 \\ \Rightarrow 3x=15 \\ \Rightarrow x=5 \end{gathered}$$

Question 9:

Solve the following equations for x:

(i) $2^{2x}-2^{x+3}+2^4=0$

(ii) $3^{2x+4}+1=2.3^{x+2}$

Answer 9:

(i)
$$\begin{gathered} 2^{2x}-2^{x+3}+2^4=0 \\ \Rightarrow {\left(2^x\right)}^2-\left(2^x\times 2^3\right)+{\left(2^2\right)}^2=0 \\ \Rightarrow {\left(2^x\right)}^2-2\times 2^x\times 2^2+{\left(2^2\right)}^2=0 \\ \Rightarrow {\left(2^x-2^2\right)}^2=0 \\ \Rightarrow 2^x-2^2=0 \\ \Rightarrow 2^x=2^2 \\ \Rightarrow x=2 \end{gathered}$$

(ii)
$$\begin{gathered} 3^{2x+4}+1=2.3^{x+2} \\ \Rightarrow {\left(3^{x+2}\right)}^2-2.3^{x+2}+1=0 \\ \Rightarrow {\left(3^{x+2}-1\right)}^2=0 \\ \Rightarrow 3^{x+2}-1=0 \\ \Rightarrow 3^{x+2}=1=3^0 \\ \Rightarrow x+2=0 \\ \Rightarrow x=-2 \end{gathered}$$

Page-2.13

Question 10:

If $49392=a^4{b}^2{c}^3$, find the values of a, b and c, where a, b and c are different positive primes.

Answer 10:

First find out the prime factorisation of 49392.

$\begin{array}{cc}2& 49392\\ 2& 24696\\ 2& 12348\\ 2& 6174\\ 3& 3087\\ 3& 1029\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}$

It can be observed that 49392 can be written as $2^4\times 3^2\times 7^3$, where 2, 3 and 7 are positive primes.
$$\begin{gathered} \therefore 49392=2^4{3}^2{7}^3=a^4{b}^2{c}^3 \\ \Rightarrow a=2, b=3, c=7 \end{gathered}$$

Question 11:

If $1176=2^a{3}^b{7}^c$, find a, b and c.

Answer 11:

First find out the prime factorisation of 1176.
$\begin{array}{cc}2& 1176\\ 2& 588\\ 2& 294\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}$

It can be observed that 1176 can be written as $2^3\times 3^1\times 7^2$.
$1176=2^3{3}^1{7}^2=2^a{3}^b{7}^c$
Hence, a = 3, b = 1 and c = 2.

Question 12:

Given $4725=3^a{5}^b{7}^c$, find
(i) the integral values of a, b and c
(ii) the value of $2^{-a}{3}^b{7}^c$

Answer 12:

(i) Given $4725=3^a{5}^b{7}^c$
First find out the prime factorisation of 4725.
$\begin{array}{cc}3& 4725\\ 3& 1575\\ 3& 525\\ 5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}$
It can be observed that 4725 can be written as $3^3\times 5^2\times 7^1$.
$\therefore 4725=3^a{5}^b{7}^c=3^3{5}^2{7}^1$
Hence, a = 3, b = 2 and c = 1.

(ii)
When a = 3, b = 2 and c = 1,
$$\begin{gathered} 2^{-a}{3}^b{7}^c \\ =2^{-3}\times 3^2\times 7^1 \\ =\frac{1}{8}\times 9\times 7 \\ =\frac{63}{8} \end{gathered}$$
Hence, the value of $2^{-a}{3}^b{7}^c$ is $\frac{63}{8}$.

Question 13:

If $a=x{y}^{p-1},b=x{y}^{q-1} \mathrm{and} c=x{y}^{r-1}$, prove that $a^{q-r}{b}^{r-p}{c}^{p-q}=1$.

Answer 13:

It is given that $a=x{y}^{p-1},b=x{y}^{q-1} \mathrm{and} c=x{y}^{r-1}$.
$$\begin{gathered} \therefore a^{q-r}{b}^{r-p}{c}^{p-q} \\ ={\left(x{y}^{p-1}\right)}^{q-r}{\left(x{y}^{q-1}\right)}^{r-p}{\left(x{y}^{r-1}\right)}^{p-q} \\ =x^{\left(q-r\right)}{y}^{\left(p-1\right)\left(q-r\right)}{x}^{\left(r-p\right)}{y}^{\left(r-p\right)\left(q-1\right)}{x}^{\left(p-q\right)}{y}^{\left(p-q\right)\left(r-1\right)} \\ =x^{\left(q-r\right)}{x}^{\left(r-p\right)}{x}^{\left(p-q\right)}{y}^{\left(p-1\right)\left(q-r\right)}{y}^{\left(r-p\right)\left(q-1\right)}{y}^{\left(p-q\right)\left(r-1\right)} \\ =x^{\left(q-r\right)+\left(r-p\right)+\left(p-q\right)}{y}^{\left(p-1\right)\left(q-r\right)+\left(r-p\right)\left(q-1\right)+\left(p-q\right)\left(r-1\right)} \\ =x^{q-r+r-p+p-q}{y}^{pq-q-pr+r+rq-r-pq+p+pr-p-qr+q} \\ =x^0{y}^0 \\ =1 \end{gathered}$$