RD Sharma 2020 solution class 9 chapter 2 Exponents of Real Numbers Exercise 2.2

Exercise 2.2

Page-2.24

Question 1:

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

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

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

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

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

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

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

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

Answer 1:

We have to simplify the following, assuming thatare positive real numbers

(i) Given

As x is positive real number then we have

Hence the simplified value of is

(ii) Given

As x and y are positive real numbers then we can write 

By using law of rational exponents we have 

Hence the simplified value of is

(iii) Given

As x and y are positive real numbers then we have 

By using law of rational exponents we have 

By using law of rational exponents we have

$$\begin{gathered} {\left(x^{\frac{-2}{3}}{y}^{{}^{\frac{-1}{2}}}\right)}^2=\frac{1}{x^{\frac{2}{3}+\frac{2}{3}}}\times \frac{1}{y^{\frac{1}{2}+\frac{1}{2}}} \\ =\frac{1}{x^{\frac{4}{3}}}\times \frac{1}{y^{\frac{2}{2}}}=\frac{1}{x^{\frac{4}{3}}}\times \frac{1}{y} \\ =\frac{1}{x^{\frac{4}{3}}y} \end{gathered}$$

Hence the simplified value of is .

 $$\begin{gathered} \left(\mathrm{iv}\right) {\left(\sqrt{x}\right)}^{-\frac{2}{3}} \sqrt{y^4} ÷ \sqrt{x{y}^{-\frac{1}{2}}} \\ ={\left(x^{\frac{1}{2}}\right)}^{-\frac{2}{3}} {\left(y^4\right)}^{\frac{1}{2}} ÷ {\left(x \times y^{-\frac{1}{2}}\right)}^{\frac{1}{2}} \\ =\frac{x^{{\displaystyle \frac{1}{2}}\times -{\displaystyle \frac{2}{3}}} \times y^{4\times {\displaystyle \frac{1}{2}}}}{x^{{\displaystyle \frac{1}{2}}} \times y^{-{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}}} \\ =\frac{x^{-{\displaystyle \frac{1}{3}}} \times y^2}{x^{{\displaystyle \frac{1}{2}}} \times y^{-{\displaystyle \frac{1}{4}}}} \\ \mathrm{by} \mathrm{using} \mathrm{the} \mathrm{law} \mathrm{of} \mathrm{rational} \mathrm{exponents}, a^m ÷ a^n = a^{m-n}, \mathrm{we} \mathrm{have} \\ x^{-\frac{1}{3}-\frac{1}{2}} \times y^{2+\frac{1}{4}} \\ =x^{-\frac{5}{6} }\times y^{\frac{9}{4}} \\ =\frac{y^{{\displaystyle \frac{9}{4}}}}{x^{{\displaystyle \frac{5}{6}}}} \\ \left(\mathrm{v}\right). \sqrt[5]{243 x^{10} y^5 z^{10}} \\ ={\left(243 \times x^{10} \times y^5 \times z^{10}\right)}^{\frac{1}{5}} \\ ={\left(243\right)}^{\frac{1}{5}} \times {\left(x^{10}\right)}^{\frac{1}{5}} \times {\left(y^5\right)}^{\frac{1}{5}} \times {\left(z^{10}\right)}^{\frac{1}{5}} \\ ={\left(3^5\right)}^{\frac{1}{5}} \times x^{10\times \frac{1}{5}} \times y^{5\times \frac{1}{5}} \times z^{10\times \frac{1}{5}} \\ =3 \times x^2 \times y \times z^2 \\ =3x^{2}y{z}^2 \\ \left(\mathrm{vi}\right) {\left(\frac{x^{-4}}{y^{-10}}\right)}^{\frac{5}{4}} \\ =\frac{{\left(x^{-4}\right)}^{{\displaystyle \frac{5}{4}}}}{{\left(y^{-10}\right)}^{{\displaystyle \frac{5}{4}}}} \\ =\frac{x^{-4\times {\displaystyle \frac{5}{4}}}}{y^{-10\times {\displaystyle \frac{5}{4}}}} \\ =\frac{x^{-5}}{y^{-{\displaystyle \frac{25}{2}}}} \\ =\frac{y^{{\displaystyle \frac{25}{2}}}}{x^5} \end{gathered}$$

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

$$\begin{gathered} {\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^5{\left(\frac{6}{7}\right)}^2 \\ ={\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^{2+2+1}{\left(\frac{6}{7}\right)}^2 \\ ={\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^2\times {\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^2\times {\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^1\times {\left(\frac{6}{7}\right)}^2 \\ =\left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)\times {\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^1\times {\left(\frac{6}{7}\right)}^2 \end{gathered}$$
$$\begin{gathered} =\left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)\times {\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^1\times {\left(\frac{6}{7}\right)}^2 \\ =\frac{16\sqrt{2}}{49\sqrt{3}} \\ =\sqrt{\frac{512}{7203}}={\left(\frac{512}{7203}\right)}^{\frac{1}{2}} \end{gathered}$$

 

Question 2:

Simplify:

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

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

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

(iv) $(0.001{)}^{1/3}$

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

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

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

Answer 2:

(1) Given

By using law of rational exponents we have

Hence the value of is
(ii) $\sqrt[5]{{\left(32\right)}^{-3}}$

$$\begin{gathered} =\sqrt[5]{{\left(\frac{1}{32}\right)}^3} \\ ={\left(\frac{1}{32}\right)}^{\frac{3}{5}} \\ ={\left(\frac{1}{{\left(2\right)}^5}\right)}^{\frac{3}{5}} \\ ={\left(\frac{1}{2}\right)}^{5\times \frac{3}{5}} \end{gathered}$$
$$\begin{gathered} ={\left(\frac{1}{2}\right)}^3 \\ =\left(\frac{1}{8}\right) \end{gathered}$$

(iii) Given

Hence the value of is

(iv) Given

The value of is

(v) Given

Hence the value of is

(vi) Given. So,

By using the law of rational exponents

Hence the value of is

(vii) Given . So,

Hence the value of is

Question 3:

Prove that:

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

(ii) $9^{3/2}-3\times 5^0-{{\left(\frac{1}{81}\right)}^{-}}^{1/2} = 15$

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

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

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

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

(vii) ${\left(\frac{64}{125}\right)}^{-2/3} + \frac{1}{{\left({\displaystyle \frac{256}{625}}\right)}^{1/4}} + \left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right) = \frac{65}{16}$

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

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

Answer 3:

(i) We have to prove that

By using rational exponent we get,

Hence,

(ii) We have to prove that. So,

Hence,

(iii) We have to prove that

Now,

Hence,

(iv) We have to prove that. So,

Let

Hence,

(v) We have to prove that

Let

 

Hence,

(vi) We have to prove that . So,

Let

Hence,

(vii) We have to prove that. So let

By taking least common factor we get 

 

Hence,

(viii) We have to prove that. So,

Let

Hence,

(ix) We have to prove that. So,

Let

Hence,

Page-2.25

Question 4:

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

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

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

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

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

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

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

Answer 4:

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

(ii)
$$\begin{gathered} {\left[\left\{\frac{x^{a\left(a-b\right)}}{x^{a\left(a+b\right)}}\right\}÷\left\{\frac{x^{b\left(b-a\right)}}{x^{b\left(b+a\right)}}\right\}\right]}^{a+b} \\ ={\left[\left\{\frac{x^{a\left(a-b\right)}}{x^{a\left(a+b\right)}}\right\}\times \left\{\frac{x^{b\left(b+a\right)}}{x^{b\left(b-a\right)}}\right\}\right]}^{a+b} \\ ={\left[\left\{\frac{x^{a^2-ab}}{x^{a^2+ab}}\right\}\times \left\{\frac{x^{b^2+ab}}{x^{b^2-ab}}\right\}\right]}^{a+b} \end{gathered}$$
$$\begin{gathered} ={\left[\left\{x^{a^2-ab-a^2-ab}\right\}\times \left\{x^{b^2+ab-b^2+ab}\right\}\right]}^{a+b} \\ ={\left[x^{-2ab}\times x^{2ab}\right]}^{a+b} \\ ={\left[x^{-2ab+2ab}\right]}^{a+b} \\ ={\left[x^0\right]}^{a+b} \\ =1 \end{gathered}$$

(iii)
$$\begin{gathered} {\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}{\left(x^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x\right)}^{\frac{1}{a-b}\times \frac{1}{a-c}}{{\left(x\right)}^{\frac{1}{b-c}\times }}^{\frac{1}{b-a}}{{\left(x\right)}^{\frac{1}{c-a}\times }}^{\frac{1}{c-b}} \\ ={\left(x\right)}^{\frac{1}{a-b}\times \frac{1}{a-c}+\frac{1}{b-c}\times \frac{1}{b-a}+\frac{1}{c-a}\times \frac{1}{c-b}} \\ ={\left(x\right)}^{\frac{\left(b-c\right)-\left(a-c\right)+\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}} \\ =x^0 \\ =1 \end{gathered}$$

(iv)
$$\begin{gathered} {\left(\frac{x^{a^2+b^2}}{x^{ab}}\right)}^{a+b}{\left(\frac{x^{b^2+c^2}}{x^{bc}}\right)}^{b+c}{\left(\frac{x^{c^2+a^2}}{x^{ac}}\right)}^{a+c} \\ ={\left(x^{a^2+b^2-ab}\right)}^{a+b}{\left(x^{b^2+c^2-bc}\right)}^{b+c}{\left(x^{c^2+a^2-ac}\right)}^{a+c} \\ =\left[x^{\left(a+b\right)\left(a^2+b^2-ab\right)}\right]\left[x^{\left(b+c\right)\left(b^2+c^2-bc\right)}\right]\left[x^{\left(a+c\right)\left(c^2+a^2-ac\right)}\right] \\ =\left(x^{a^3+b^3}\right)\left(x^{b^3+c^3}\right)\left(x^{a^3+c^3}\right) \\ =x^{2\left(a^3+b^3+c^3\right)} \end{gathered}$$

(v)
$$\begin{gathered} {\left(x^{a-b}\right)}^{a+b}{\left(x^{b-c}\right)}^{b+c}{\left(x^{c-a}\right)}^{c+a} \\ =\left[x^{\left(a-b\right)\left(a+b\right)}\right]\left[x^{\left(b-c\right)\left(b+c\right)}\right]\left[x^{\left(c-a\right)\left(c+a\right)}\right] \\ =x^{\left(a^2-b^2\right)}{x}^{\left(b^2-c^2\right)}{x}^{\left(c^2-a^2\right)} \\ =x^{a^2-b^2+b^2-c^2+c^2-a^2} \\ =x^0 \\ =1 \end{gathered}$$

(vi)
 $$\begin{gathered} {\left\{{\left(x^{a-a^{-1}}\right)}^{\frac{1}{a-1}}\right\}}^{\frac{a}{a+1}} \\ =\left\{{\left(x^{a-\frac{1}{a}}\right)}^{\frac{1}{a-1}\times \frac{a}{a+1}}\right\} \\ ={\left\{x^{\frac{a^2-1}{a}}\right\}}^{\frac{a}{a^2-1}} \\ =x^{\frac{a^2-1}{a}\times \frac{a}{a^2-1}} \\ =x^1 \\ =x \end{gathered}$$

(vii)
$$\begin{gathered} {\left(\frac{a^{x+1}}{a^{y+1}}\right)}^{x+y}{\left(\frac{a^{y+2}}{a^{z+2}}\right)}^{y+z}{\left(\frac{a^{z+3}}{a^{x+3}}\right)}^{z+x} \\ ={\left(a^{x+1-y-1}\right)}^{x+y}{\left(a^{y+2-z-2}\right)}^{y+z}{\left(a^{z+3-x-3}\right)}^{z+x} \\ ={\left(a^{x-y}\right)}^{x+y}{\left(a^{y-z}\right)}^{y+z}{\left(a^{z-x}\right)}^{z+x} \\ =a^{x^2-y^2+y^2-z^2+z^2-x^2} \\ =a^0 \\ =1 \end{gathered}$$

(viii)
$$\begin{gathered} {\left(\frac{3^a}{3^b}\right)}^{a+b}{\left(\frac{3^b}{3^c}\right)}^{b+c}{\left(\frac{3^c}{3^a}\right)}^{c+a} \\ ={\left(3^{a-b}\right)}^{a+b}{\left(3^{b-c}\right)}^{b+c}{\left(3^{c-a}\right)}^{c+a} \\ =3^{a^2-b^2+b^2-c^2+c^2-a^2} \\ =3^0 \\ =1 \end{gathered}$$

Question 5:

Question

Answer 5:

Let $2^x=3^y={12}^z=k$
$\Rightarrow 2=k^{\frac{1}{x}},3=k^{\frac{1}{y}},12=k^{\frac{1}{z}}$
Now,
$$\begin{gathered} 12=k^{\frac{1}{z}} \\ \Rightarrow 2^2\times 3=k^{\frac{1}{z}} \\ \Rightarrow {\left(k^{\frac{1}{x}}\right)}^2\times k^{\frac{1}{y}}=k^{\frac{1}{z}} \\ \Rightarrow k^{\frac{2}{x}+\frac{1}{y}}=k^{\frac{1}{z}} \\ \Rightarrow \frac{2}{x}+\frac{1}{y}=\frac{1}{z} \end{gathered}$$

Question 6:

Question

Answer 6:

 Let $2^x=3^y=6^{-z}=k$
$\Rightarrow 2=k^{\frac{1}{x}}, 3=k^{\frac{1}{y}},6=k^{\frac{1}{-z}}$
Now,
$$\begin{gathered} 6=2\times 3=k^{\frac{1}{-z}} \\ \Rightarrow k^{\frac{1}{x}}\times k^{\frac{1}{y}}=k^{\frac{1}{-z}} \\ \Rightarrow k^{\frac{1}{x}+\frac{1}{y}}=k^{\frac{1}{-z}} \\ \Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{-z} \\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \end{gathered}$$

Question 7:

Question

Answer 7:

Let $a^x=b^y=c^z=k$
So, $a=k^{\frac{1}{x}},b=k^{\frac{1}{y}},c=k^{\frac{1}{z}}$
Thus, 
$$\begin{gathered} b^2=ac \\ \Rightarrow {\left(k^{\frac{1}{y}}\right)}^2=\left(k^{\frac{1}{x}}\right)\left(k^{\frac{1}{z}}\right) \\ \Rightarrow k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}} \\ \Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{2}{y}=\frac{z+x}{xz} \\ \Rightarrow y=\frac{2zx}{z+x} \end{gathered}$$

Page-2.26

Question 8:

Question

Answer 8:

Let $3^x=5^y={\left(75\right)}^z=k$
$$\begin{gathered} \Rightarrow 3=k^{\frac{1}{x}}, 5=k^{\frac{1}{y}}, 75=k^{\frac{1}{z}} \\ \Rightarrow 5^2\times 3=k^{\frac{1}{z}} \\ \Rightarrow {\left(k^{\frac{1}{y}}\right)}^2\times k^{\frac{1}{x}}=k^{\frac{1}{z}} \\ \Rightarrow k^{\frac{2}{y}}\times k^{\frac{1}{x}}=k^{\frac{1}{z}} \end{gathered}$$
$$\begin{gathered} \Rightarrow k^{\frac{2}{y}+\frac{1}{x}}=k^{\frac{1}{z}} \\ \Rightarrow \frac{2}{y}+\frac{1}{x}=\frac{1}{z} \\ \Rightarrow \frac{2x+y}{xy}=\frac{1}{z} \\ \Rightarrow z=\frac{xy}{2x+y} \end{gathered}$$

Question 9:

If ${27}^x =\frac{9}{3^x},$ find x.

Answer 9:

We are given. We have to find the value of

Since

By using the law of exponents we get, 

On equating the exponents we get,

Hence,

Question 10:

Find the values of x in each of the following:

(i) $2^{5x}÷2x =\sqrt[5]{2^{20}}$

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

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

(iv) $5^{x-2}\times 3^{2x-3} =135$

(v) $2^{x-7}\times 5^{x-4}=1250$

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

(vii) $5^{2x+3}=1$

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

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

Answer 10:

From the following we have to find the value of x

(i) Given

By using rational exponents

On equating the exponents we get,

The value of x is

(ii) Given

On equating the exponents 

Hence the value of x is

(iii) Given

Comparing exponents we have,

Hence the value of x is

(iv) Given

On equating the exponents of 5 and 3 we get,

And,

The value of x is

(v) Given

On equating the exponents we get 

And, 

Hence the value of x is

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

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

On comparing we get, 
$$\begin{gathered} \frac{4x+1}{3}=-5 \\ \Rightarrow 4x+1=-15 \\ \Rightarrow 4x=-16 \\ \Rightarrow x=-4 \end{gathered}$$

(vii) $5^{2x+3}=1$

$$\begin{gathered} 5^{2x+3}=5^0 \\ \Rightarrow 2x+3=0 \\ \Rightarrow x=\frac{-3}{2} \end{gathered}$$

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

$$\begin{gathered} {\left(13\right)}^{\sqrt{x}}={\left(2^2\right)}^4-3^4-6 \\ \Rightarrow {\left(13\right)}^{\sqrt{x}}=2^8-3^4-6 \\ \Rightarrow {\left(13\right)}^{\sqrt{x}}=256-81-6 \\ \Rightarrow {\left(13\right)}^{\sqrt{x}}=169 \\ \Rightarrow {\left(13\right)}^{\sqrt{x}}={\left(13\right)}^2 \end{gathered}$$

On comparing we get, 
$$\begin{gathered} \sqrt{x}=2 \\ \mathrm{on} \mathrm{squaring} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get}, \\ \mathrm{x}=4 \end{gathered}$$

(ix) ${\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}$
$$\begin{gathered} {\left(\sqrt{\frac{3}{5}}\right)}^{x+1}={\left(\frac{5}{3}\right)}^3 \\ \Rightarrow {\left(\frac{3}{5}\right)}^{\frac{x+1}{2}}={\left(\frac{3}{5}\right)}^{-3} \end{gathered}$$

On comparing we get, 

$$\begin{gathered} \frac{x+1}{2}=-3 \\ \Rightarrow x+1=-6 \\ \Rightarrow x=-7 \end{gathered}$$

Question 11:

Question

Answer 11:

 $x=2^{\frac{1}{3}}+2^{\frac{2}{3}}$
Cubing on both sides, we get
$$\begin{gathered} x^3={\left(2^{\frac{1}{3}}+2^{\frac{2}{3}}\right)}^3 \\ \Rightarrow x^3={\left(2^{\frac{1}{3}}\right)}^3+{\left(2^{\frac{2}{3}}\right)}^3+3\times 2^{\frac{1}{3}}\times 2^{\frac{2}{3}}\left(2^{\frac{1}{3}}+2^{\frac{2}{3}}\right) \\ \Rightarrow x^3=2+2^2+3\times 2^{\frac{1+2}{3}}\left(x\right) \\ \Rightarrow x^3=2+4+3\times 2x \\ \Rightarrow x^3-6x=6 \end{gathered}$$

Question 12:

Question

Answer 12:

$9^{x+2}=240+9^x$
$$\begin{gathered} \Rightarrow 9^x\times 9^2=240+9^x \\ \Rightarrow 9^x\left(9^2-1\right)=240 \\ \Rightarrow 9^x\left(81-1\right)=240 \\ \Rightarrow 9^x\times 80=240 \\ \Rightarrow 9^x=3 \\ \Rightarrow 3^{2x}=3^1 \\ \Rightarrow 2x=1 \\ \Rightarrow x=\frac{1}{2} \end{gathered}$$
 $\therefore {\left(8x\right)}^x={\left(8\times \frac{1}{2}\right)}^{\frac{1}{2}}={\left(4\right)}^{\frac{1}{2}}=2$

Question 13:

Question

Answer 13:

 $3^{x+1}=9^{x-2}$
$$\begin{gathered} \Rightarrow 3^x\times 3=\frac{9^x}{9^2} \\ \Rightarrow 3^x\times 3=\frac{{\left(3^2\right)}^x}{{\left(3^2\right)}^2}=\frac{3^{2x}}{3^4} \\ \Rightarrow 3^4\times 3=\frac{3^{2x}}{3^x} \\ \Rightarrow 3^5=3^{2x-x} \end{gathered}$$
$\Rightarrow 3^5=3^x$
Comparing both sides, we get
x = 5
So, $2^{1+x}=2^{1+5}=2^6=64$

Question 14:

If $3^{4x}={\left(81\right)}^{-1}$ and ${10}^{\frac{1}{y}}=0.0001$, find the value of $2^{-x+4y}$.

Answer 14:

It is given that $3^{4x}={\left(81\right)}^{-1}$ and ${10}^{\frac{1}{y}}=0.0001$.
Now,
$$\begin{gathered} 3^{4x}={\left(81\right)}^{-1} \\ \Rightarrow 3^{4x}={\left(3^4\right)}^{-1} \\ \Rightarrow {\left(3^x\right)}^4={\left(3^{-1}\right)}^4 \\ \Rightarrow x=-1 \end{gathered}$$
And,
$$\begin{gathered} {10}^{\frac{1}{y}}=0.0001 \\ \Rightarrow {10}^{\frac{1}{y}}=\frac{1}{10000} \\ \Rightarrow {10}^{\frac{1}{y}}={\left(\frac{1}{10}\right)}^4 \\ \Rightarrow {10}^{\frac{1}{y}}={\left(10\right)}^{-4} \\ \Rightarrow \frac{1}{y}=-4 \\ \Rightarrow y=-\frac{1}{4} \end{gathered}$$
Therefore, the value of $2^{-x+4y}$ is $2^{1+4\left(-\frac{1}{4}\right)}=2^0=1$.

Question 15:

If $5^{3x}=125 \text{and }{10}^y=0.001$, find x and y.

Answer 15:

It is given that $5^{3x}=125 \text{and }{10}^y=0.001$.
Now,
$$\begin{gathered} 5^{3x}=125 \\ \Rightarrow 5^{3x}=5^3 \\ \Rightarrow 3x=3 \\ \Rightarrow x=1 \end{gathered}$$
And,
$$\begin{gathered} {10}^y=0.001 \\ \Rightarrow {10}^y=\frac{1}{1000} \\ \Rightarrow {10}^y={10}^{-3} \\ \Rightarrow y=-3 \end{gathered}$$
Hence, the values of x and y are 1 and −3, respectively.

Question 16:

Solve the following equations:
(i) $3^{x+1}=27\times 3^4$
(ii) $4^{2x}={\left(\sqrt[3]{16}\right)}^{-\frac{6}{y}}={\left(\sqrt{8}\right)}^2$
(iii) $3^{x-1}\times 5^{2y-3}=225$
(iv) $8^{x+1}={16}^{y+2} \mathrm{and}, {\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{4}\right)}^{3y}$
(v) $4^{x-1}\times {\left(0.5\right)}^{3-2x}={\left(\frac{1}{8}\right)}^x$

(vi) $\sqrt{\frac{a}{b}}={\left(\frac{b}{a}\right)}^{1-2x}$, where a and b are distinct primes

Answer 16:

(i)
$$\begin{gathered} 3^{x+1}=27\times 3^4 \\ \Rightarrow 3^{x+1}=3^3\times 3^4 \\ \Rightarrow 3^{x+1}=3^{3+4} \\ \Rightarrow 3^{x+1}=3^7 \\ \Rightarrow x+1=7 \\ \Rightarrow x=6 \end{gathered}$$

(ii)
$$\begin{gathered} 4^{2x}={\left(\sqrt[3]{16}\right)}^{-\frac{6}{y}}={\left(\sqrt{8}\right)}^2 \\ \Rightarrow 4^{2x}={\left(\sqrt{8}\right)}^2 \mathrm{and} {\left(\sqrt[3]{16}\right)}^{-\frac{6}{y}}={\left(\sqrt{8}\right)}^2 \\ \Rightarrow 4^{2x}=\left(8^{\frac{1}{2}\times 2}\right) \mathrm{and} \left({16}^{\frac{1}{3}\times -\frac{6}{y}}\right)=\left(8^{\frac{1}{2}\times 2}\right) \\ \Rightarrow 4^{2x}=8 \mathrm{and} \left({16}^{-\frac{2}{y}}\right)=8 \\ \Rightarrow 2^{4x}=2^3 \mathrm{and} \left(2^{-\frac{8}{y}}\right)=2^3 \\ \Rightarrow 4x=3 \mathrm{and} -\frac{8}{y}=3 \\ \Rightarrow x=\frac{3}{4} \mathrm{and} y=-\frac{8}{3} \end{gathered}$$

(iii)
$$\begin{gathered} 3^{x-1}\times 5^{2y-3}=225 \\ \Rightarrow 3^{x-1}\times 5^{2y-3}=3\times 3\times 5\times 5 \\ \Rightarrow 3^{x-1}\times 5^{2y-3}=3^2\times 5^2 \\ \Rightarrow x-1=2 \mathrm{and} 2y-3=2 \\ \Rightarrow x=3 \mathrm{and} y=\frac{5}{2} \end{gathered}$$

(iv)
$$\begin{gathered} 8^{x+1}={16}^{y+2} \mathrm{and}, {\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{4}\right)}^{3y} \\ \Rightarrow {\left(2^3\right)}^{x+1}={\left(2^4\right)}^{y+2} \mathrm{and} {\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{2^2}\right)}^{3y} \\ \Rightarrow {\left(2\right)}^{3x+3}={\left(2\right)}^{4y+8} \mathrm{and} {\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{2}\right)}^{6y} \\ \Rightarrow 3x+3=4y+8 \mathrm{and} 3+x=6y \end{gathered}$$
Now,
$3+x=6y\Rightarrow x=6y-3$
Putting x = 6y − 3 in $3x-4y=5$, we get
$$\begin{gathered} 3\left(6y-3\right)-4y=5 \\ \Rightarrow 18y-9-4y=5 \\ \Rightarrow 14y=14 \\ \Rightarrow y=1 \end{gathered}$$
Putting y = 1 in $x=6y-3$, we get
$x=6\times 1-3=3$

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

(vi)
$$\begin{gathered} \sqrt{\frac{a}{b}}={\left(\frac{b}{a}\right)}^{1-2x} \\ \Rightarrow {\left(\frac{a}{b}\right)}^{\frac{1}{2}}={\left(\frac{a}{b}\right)}^{-\left(1-2x\right)} \\ \Rightarrow \frac{1}{2}=-\left(1-2x\right) \\ \Rightarrow \frac{1}{2}=2x-1 \\ \Rightarrow \frac{3}{2}=2x \\ \Rightarrow x=\frac{3}{4} \end{gathered}$$

Question 17:

If a and b are distinct primes such that $\sqrt[3]{a^6{b}^{-4}}=a^x{b}^{2y}$, find x and y.

Answer 17:

Given: $\sqrt[3]{a^6{b}^{-4}}=a^x{b}^{2y}$
$$\begin{gathered} \sqrt[3]{a^6{b}^{-4}}=a^x{b}^{2y} \\ \Rightarrow {\left(a^6{b}^{-4}\right)}^{\frac{1}{3}}=a^x{b}^{2y} \\ \Rightarrow a^{6\times \frac{1}{3}}{b}^{-4\times \frac{1}{3}}=a^x{b}^{2y} \\ \Rightarrow a^2{b}^{-\frac{4}{3}}=a^x{b}^{2y} \\ \Rightarrow x=2 \mathrm{and} y=-\frac{2}{3} \end{gathered}$$

Question 18:

If a and b are different positive primes such that

(i) ${\left(\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right)}^7÷\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)=a^x{b}^y$, find x and y.

(ii) ${\left(a+b\right)}^{-1}\left(a^{-1}+b^{-1}\right)=a^x{b}^y$, find x + y + 2.

Answer 18:

(i)
$$\begin{gathered} {\left(\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right)}^7÷\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)=a^x{b}^y \\ \Rightarrow \left(\frac{a^{-7}{b}^{14}}{a^{14}{b}^{-28}}\right)÷\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)=a^x{b}^y \\ \Rightarrow \left(a^{-7-14}{b}^{14+28}\right)÷\left(a^{3+2}{b}^{-5-3}\right)=a^x{b}^y \\ \Rightarrow \left(a^{-21}{b}^{42}\right)÷\left(a^5{b}^{-8}\right)=a^x{b}^y \\ \Rightarrow a^{-21-5}{b}^{42+8}=a^x{b}^y \\ \Rightarrow a^{-26}{b}^{50}=a^x{b}^y \\ \Rightarrow x=-26 \mathrm{and} y=50 \end{gathered}$$

(ii)
$$\begin{gathered} {\left(a+b\right)}^{-1}\left(a^{-1}+b^{-1}\right)=a^x{b}^y \\ \Rightarrow \frac{1}{\left(a+b\right)}\left(\frac{1}{a}+\frac{1}{b}\right)=a^x{b}^y \\ \Rightarrow \frac{1}{\left(a+b\right)}\left(\frac{a+b}{ab}\right)=a^x{b}^y \\ \Rightarrow \left(\frac{1}{ab}\right)=a^x{b}^y \\ \Rightarrow {\left(ab\right)}^{-1}=a^x{b}^y \\ \Rightarrow a^{-1}{b}^{\mathbf{-}\mathbf{1}}=a^x{b}^y \\ \Rightarrow x=-1 \mathrm{and} y=-1 \end{gathered}$$
Therefore, the value of x + y + 2 is −1 −1 + 2 = 0.

Question 19:

If $2^x\times 3^y\times 5^z=2160$, find x , y and z. Hence, compute the value of $3^x\times 2^{-y}\times 5^{-z}$.

Answer 19:

Given: $2^x\times 3^y\times 5^z=2160$
First, find out the prime factorisation of 2160.
$\begin{array}{cc}2& 2160\\ 2& 1080\\ 2& 540\\ 2& 270\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$
It can be observed that 2160 can be written as $2^4\times 3^3\times 5^1$.
Also,
$$\begin{gathered} 2^x\times 3^y\times 5^z=2^4\times 3^3\times 5^1 \\ \Rightarrow x=4, y=3, z=1 \end{gathered}$$
Therefore, the value of $3^x\times 2^{-y}\times 5^{-z}$ is $3^4\times 2^{-3}\times 5^{-1}=81\times \frac{1}{8}\times \frac{1}{5}=\frac{81}{40}$.

Question 20:

If $1176=2^a\times 3^b\times 7^c$, find the values of a, b and c. Hence, compute the value of $2^a\times 3^b\times 7^{-c}$ as a fraction.

Answer 20:

First find 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 $2^3\times 3^1\times 7^2$.
$1176=2^a{3}^b{7}^c=2^3{3}^1{7}^2$
So, a = 3, b = 1 and c = 2.
Therefore, the value of $2^a\times 3^b\times 7^{-c}$  is $2^3\times 3^1\times 7^{-2}=8\times 3\times \frac{1}{49}=\frac{24}{49}$

Page-2.27

Question 21:

Simplify:

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

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

Answer 21:

(i)
$$\begin{gathered} {\left(\frac{x^{a+b}}{x^c}\right)}^{a-b}{\left(\frac{x^{b+c}}{x^a}\right)}^{b-c}{\left(\frac{x^{c+a}}{x^b}\right)}^{c-a} \\ =\left(\frac{x^{\left(a+b\right)\left(a-b\right)}}{x^{c\left(a-b\right)}}\right)\left(\frac{x^{\left(b+c\right)\left(b-c\right)}}{x^{a\left(b-c\right)}}\right)\left(\frac{x^{\left(c+a\right)\left(c-a\right)}}{x^{b\left(c-a\right)}}\right) \\ =\left(\frac{x^{\left(a+b\right)\left(a-b\right)}}{x^{ca-bc}}\right)\left(\frac{x^{\left(b+c\right)\left(b-c\right)}}{x^{ab-ac}}\right)\left(\frac{x^{\left(c+a\right)\left(c-a\right)}}{x^{bc-ab}}\right) \\ =\left(\frac{x^{\left(a^2-b^2\right)}{x}^{\left(b^2-c^2\right)}{x}^{\left(c^2-a^2\right)}}{x^{ca-bc+ab-ac+bc-ab}}\right) \\ =\frac{x^{a^2-b^2+b^2-c^2+c^2-a^2}}{x^{ca-bc+ab-ac+bc-ab}} \\ =\frac{x^0}{x^0} \\ =1 \end{gathered}$$

(ii)
$$\begin{gathered} \sqrt[lm]{\frac{x^l}{x^m}}\times \sqrt[mn]{\frac{x^m}{x^n}}\times \sqrt[nl]{\frac{x^n}{x^l}} \\ ={\left(\frac{x^l}{x^m}\right)}^{\frac{1}{ml}}\times {\left(\frac{x^m}{x^n}\right)}^{\frac{1}{mn}}\times {\left(\frac{x^n}{x^l}\right)}^{\frac{1}{nl}} \\ ={\left(x^{l-m}\right)}^{\frac{1}{ml}}\times {\left(x^{m-n}\right)}^{\frac{1}{mn}}\times {\left(x^{n-l}\right)}^{\frac{1}{nl}} \\ =x^{\frac{l-m}{ml}}\times x^{\frac{m-n}{mn}}\times x^{\frac{n-l}{nl}} \\ =x^{\frac{l-m}{ml}+\frac{m-n}{mn}+\frac{n-l}{nl}} \\ =x^{\frac{ln-mn+lm-nl+nm-lm}{nml}} \\ =x^0 \\ =1 \end{gathered}$$

Question 22:

Show that: $\frac{{\left(a+{\displaystyle \frac{1}{b}}\right)}^m\times {\left(a-{\displaystyle \frac{1}{b}}\right)}^n}{{\left(b+{\displaystyle \frac{1}{a}}\right)}^m\times {\left(b-{\displaystyle \frac{1}{a}}\right)}^n}={\left(\frac{a}{b}\right)}^{m+n}$

Answer 22:

$$\begin{gathered} \frac{{\left(a+{\displaystyle \frac{1}{b}}\right)}^m\times {\left(a-{\displaystyle \frac{1}{b}}\right)}^n}{{\left(b+{\displaystyle \frac{1}{a}}\right)}^m\times {\left(b-{\displaystyle \frac{1}{a}}\right)}^n} \\ =\frac{{\left({\displaystyle \frac{ab+1}{b}}\right)}^m\times {\left({\displaystyle \frac{ab-1}{b}}\right)}^n}{{\left({\displaystyle \frac{ab+1}{a}}\right)}^m\times {\left({\displaystyle \frac{ab-1}{a}}\right)}^n} \\ ={\left(\frac{{\displaystyle \frac{ab+1}{b}}}{{\displaystyle \frac{ab+1}{a}}}\right)}^m\times {\left(\frac{{\displaystyle \frac{ab-1}{b}}}{{\displaystyle \frac{ab-1}{a}}}\right)}^n \\ ={\left(\frac{ab+1}{b}\times \frac{{\displaystyle a}}{{\displaystyle ab+1}}\right)}^m\times {\left(\frac{ab-1}{b}\times \frac{{\displaystyle a}}{{\displaystyle ab-1}}\right)}^n \\ ={\left(\frac{{\displaystyle a}}{{\displaystyle b}}\right)}^m\times {\left(\frac{{\displaystyle a}}{{\displaystyle b}}\right)}^n \\ ={\left(\frac{{\displaystyle a}}{{\displaystyle b}}\right)}^m\times {\left(\frac{{\displaystyle a}}{{\displaystyle b}}\right)}^n \\ ={\left(\frac{a}{b}\right)}^{m+n} \end{gathered}$$

Question 23:

(i) If $a=x^{m+n}{y}^l, b=x^{n+l}{y}^m \mathrm{and} c=x^{l+m}{y}^n$, prove that $a^{m-n}{b}^{n-l}{c}^{l-m}=1$.

(ii) If $x=a^{m+n}, y=a^{n+l} \mathrm{and} z=a^{l+m}$, prove that $x^m{y}^n{z}^l=x^n{y}^l{z}^m$.

Answer 23:

(i) Given: $a=x^{m+n}{y}^l, b=x^{n+l}{y}^m \mathrm{and} c=x^{l+m}{y}^n$
Putting the values of a, b and c in $a^{m-n}{b}^{n-l}{c}^{l-m}$, we get
$$\begin{gathered} a^{m-n}{b}^{n-l}{c}^{l-m} \\ ={\left(x^{m+n}{y}^l\right)}^{m-n}{\left(x^{n+l}{y}^m\right)}^{n-l}{\left(x^{l+m}{y}^n\right)}^{l-m} \\ =\left[x^{\left(m+n\right)\left(m-n\right)}{y}^{l\left(m-n\right)}\right]\left[x^{\left(n+l\right)\left(n-l\right)}{y}^{m\left(n-l\right)}\right]\left[x^{\left(l+m\right)\left(l-m\right)}{y}^{n\left(l-m\right)}\right] \\ =x^{\left(m^2-n^2\right)}{x}^{\left(n^2-l^2\right)}{x}^{\left(l^2-m^2\right)}{y}^{lm-ln}{y}^{mn-ml}{y}^{nl-nm} \\ =x^{m^2-n^2+n^2-l^2+l^2-m^2}{y}^{lm-ln+mn-ml+nl-nm} \\ =x^0{y}^0 \\ =1 \end{gathered}$$

(ii) Given: $x=a^{m+n}, y=a^{n+l} \mathrm{and} z=a^{l+m}$
Putting the values of x, y and z in $x^m{y}^n{z}^l$, we get
$$\begin{gathered} x^m{y}^n{z}^l \\ ={\left(a^{m+n}\right)}^m{\left(a^{n+l}\right)}^n{\left(a^{l+m}\right)}^l \\ =\left(a^{m^2+nm}\right)\left(a^{n^2+\mathrm{l}n}\right)\left(a^{l^2+lm}\right) \\ =a^{m^2+n^2+l^2+nm+ln+lm} \end{gathered}$$
Putting the values of x, y and z in $x^n{y}^l{z}^m$, we get
$$\begin{gathered} x^n{y}^l{z}^m \\ ={\left(a^{m+n}\right)}^n{\left(a^{n+l} \right)}^l{\left(a^{l+m}\right)}^m \\ =\left(a^{mn+n^2}\right)\left(a^{nl+l^2} \right)\left(a^{lm+m^2}\right) \\ =a^{mn+n^2+nl+l^2+lm+m^2} \end{gathered}$$
So, $x^m{y}^n{z}^l$ = $x^n{y}^l{z}^m$