RD Sharma solution class 7 chapter 6 Exponents Objective Type Question

Objective Type Question

Page-6.32

Question 1:

Mark the correct alternative in the following question:

$$\begin{gathered} {\left(6^{-1}-8^{-1}\right)}^{-1}= \\ \left(\mathrm{a}\right) \frac{1}{24} \left(\mathrm{b}\right) 24 \left(\mathrm{c}\right) -24 \left(\mathrm{d}\right) -\frac{1}{24} \end{gathered}$$

Answer 1:

$$\begin{gathered} {\left(6^{-1}-8^{-1}\right)}^{-1} \\ ={\left(\frac{1}{6}-\frac{1}{8}\right)}^{-1} \left(\mathrm{As}, x^{-1}=\frac{1}{x}\right) \\ ={\left(\frac{4-3}{24}\right)}^{-1} \\ ={\left(\frac{1}{24}\right)}^{-1} \\ =\frac{24}{1} \left(\mathrm{As}, x^{-1}=\frac{1}{x}\right) \\ =24 \end{gathered}$$

Hence, the correct alternative is option (b).

Question 2:

Mark the correct alternative in the following question:

$$\begin{gathered} 2^{3^2}= \\ \left(\mathrm{a}\right) 64 \left(\mathrm{b}\right) 32 \left(\mathrm{c}\right) 256 \left(\mathrm{d}\right) 512 \end{gathered}$$

Answer 2:

Since, $2^{3^2}=2^9=512$

Hence, the correct alternative is option (d).

Question 3:

Mark the correct alternative in the following question:

$$\begin{gathered} {\left(3^{-1}\times 5^{-1}\right)}^{-1}= \\ \left(\mathrm{a}\right) \frac{1}{15} \left(\mathrm{b}\right) -\frac{1}{15} \left(\mathrm{c}\right) 15 \left(\mathrm{d}\right) -15 \end{gathered}$$

Answer 3:

$$\begin{gathered} {\left(3^{-1}\times 5^{-1}\right)}^{-1} \\ ={\left(\frac{1}{3}\times \frac{1}{5}\right)}^{-1} \left(\mathrm{As}, x^{-1}=\frac{1}{x}\right) \\ ={\left(\frac{1}{15}\right)}^{-1} \\ =\frac{15}{1} \left(\mathrm{As}, x^{-1}=\frac{1}{x}\right) \\ =15 \end{gathered}$$

Hence, the correct alternative is option (c).

Question 4:

Mark the correct alternative in the folowing question:

$$\begin{gathered} {\left(-\frac{3}{5}\right)}^{-1}= \\ \left(\mathrm{a}\right) \frac{3}{5} \left(\mathrm{b}\right) \frac{5}{3} \left(\mathrm{c}\right) -\frac{5}{3} \left(\mathrm{d}\right) -\frac{3}{5} \end{gathered}$$

Answer 4:

Since, ${\left(-\frac{3}{5}\right)}^{-1}=-\frac{5}{3} \left(\mathrm{As}, x^{-1}=\frac{1}{x}\right)$

Hence, the correct alternative is option (c).

Question 5:

Mark the correct alternative in the folowing question:

$$\begin{gathered} {\left(-1\right)}^{301}+{\left(-1\right)}^{302}+{\left(-1\right)}^{303}+...+{\left(-1\right)}^{400} \\ \left(\mathrm{a}\right) 1 \left(\mathrm{b}\right) 101 \left(\mathrm{c}\right) 100 \left(\mathrm{d}\right) 0 \end{gathered}$$

Answer 5:

Since,

$$\begin{gathered} {\left(-1\right)}^{301}+{\left(-1\right)}^{302}+{\left(-1\right)}^{303}+...+{\left(-1\right)}^{400} \\ =\left(-1\right)+\left(1\right)+\left(-1\right)+...+\left(1\right) \left[\mathrm{As}, {\left(-1\right)}^{\mathrm{odd}}=-1 \mathrm{and} {\left(-1\right)}^{\mathrm{even}}=1\right] \\ =-50+50 \left[\mathrm{As}, \mathrm{there} \mathrm{are} 50 \left(-1\right)\text{'}\mathrm{s} \mathrm{and} 50 \left(1\right)\text{'}\mathrm{s}\right] \\ =0 \end{gathered}$$

Hence, the correct alternative is option is (d).

Question 6:

Mark the correct alternative in the folowing question:

$$\begin{gathered} \mathrm{If} a=25, \mathrm{then} a^{{25}^0}+a^{0^{25}}= \\ \left(\mathrm{a}\right) 25 \left(\mathrm{b}\right) 26 \left(\mathrm{c}\right) 24 \left(\mathrm{d}\right) 0 \end{gathered}$$

Answer 6:

Since,

$$\begin{gathered} a^{{25}^0}+a^{0^{25}} \\ =a^1+a^0 \left(\mathrm{As}, {25}^0=1 \mathrm{and} 0^{25}=0\right) \\ =a+1 \left(\mathrm{As}, a^0=1\right) \\ =25+1 \\ =26 \end{gathered}$$

Hence, the correct alternative is option (b).

Question 7:

Mark the correct alternative in the following question:

$$\begin{gathered} \left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}= \\ \left(\mathrm{a}\right) \frac{19}{64} \left(\mathrm{b}\right) \frac{64}{19} \left(\mathrm{c}\right) \frac{27}{16} \left(\mathrm{d}\right) \frac{-19}{64} \end{gathered}$$

Answer 7:

Since,

$$\begin{gathered} \left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3} \\ =\left\{{\left(\frac{3}{1}\right)}^3-{\left(\frac{2}{1}\right)}^3\right\}÷{\left(\frac{4}{1}\right)}^3 \left(\mathrm{As}, x^{-1}=\frac{1}{x}\right) \\ =\left\{3^3-2^3\right\}÷4^3 \\ =\left\{27-8\right\}÷64 \\ =19÷64 \\ =\frac{19}{64} \end{gathered}$$

Hence, the correct alternative is option (a).

Question 8:

Mark the correct alternative in the following question:

$$\begin{gathered} {\left({25}^2-{15}^2\right)}^{\frac{3}{2}}= \\ \left(\mathrm{a}\right) 4000 \left(\mathrm{b}\right) 8000 \left(\mathrm{c}\right) 3125 \left(\mathrm{d}\right) 1024 \end{gathered}$$

Answer 8:

Since,

$$\begin{gathered} {\left({25}^2-{15}^2\right)}^{\frac{3}{2}} \\ ={\left(625-225\right)}^{\frac{3}{2}} \\ ={\left(400\right)}^{\frac{3}{2}} \\ ={\left({20}^2\right)}^{\frac{3}{2}} \\ ={20}^{2\times \frac{3}{2}} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ ={20}^3 \\ =8000 \end{gathered}$$

Hence, the correct alternative is option (b).

Question 9:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} {\left(\frac{5}{3}\right)}^{-5}\times {\left(\frac{5}{3}\right)}^{11}={\left(\frac{5}{3}\right)}^{8x}, \mathrm{then} x=? \\ \left(\mathrm{a}\right) -\frac{1}{2} \left(\mathrm{b}\right) -\frac{3}{4} \left(\mathrm{c}\right) \frac{3}{4} \left(\mathrm{d}\right) \frac{4}{3} \end{gathered}$$

Answer 9:

Since,

$$\begin{gathered} {\left(\frac{5}{3}\right)}^{-5}\times {\left(\frac{5}{3}\right)}^{11}={\left(\frac{5}{3}\right)}^{8x} \\ \Rightarrow {\left(\frac{5}{3}\right)}^{-5+11}={\left(\frac{5}{3}\right)}^{8x} \left(\mathrm{As}, x^m{\times }^n=x^{m+n}\right) \\ \Rightarrow {\left(\frac{5}{3}\right)}^6={\left(\frac{5}{3}\right)}^{8x} \\ \mathrm{Comparing} \mathrm{the} \mathrm{exponent} \mathrm{of} \mathrm{both} \mathrm{the} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ 8x=6 \\ \Rightarrow x=\frac{6}{8} \\ \therefore x=\frac{3}{4} \end{gathered}$$

Hence, the correct alternative is option (c).

Question 10:

Mark the correct alternative in the following question:

$$\begin{gathered} {\left[{\left\{{\left(-\frac{1}{3}\right)}^2\right\}}^{-2}\right]}^{-1}= \\ \left(\mathrm{a}\right) \frac{1}{81} \left(\mathrm{b}\right) \frac{1}{9} \left(\mathrm{c}\right) -\frac{1}{81} \left(\mathrm{d}\right) -\frac{1}{9} \end{gathered}$$

Answer 10:

$$\begin{gathered} {\left[{\left\{{\left(-\frac{1}{3}\right)}^2\right\}}^{-2}\right]}^{-1} \\ =\left[{\left\{{\left(\frac{-1}{3}\right)}^2\right\}}^{\left(-2\right)\times \left(-1\right)}\right] \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ ={\left(\frac{-1}{3}\right)}^{2\times 2} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ ={\left(\frac{-1}{3}\right)}^4 \\ =\frac{{\left(-1\right)}^4}{3^4} \left[\mathrm{As}, {\left(\frac{x}{y}\right)}^m=\frac{x^m}{y^m}\right] \\ =\frac{1}{81} \end{gathered}$$

Hence, the correct alternative is option (a).

Question 11:

Mark the correct alternative in the following question:

$$\begin{gathered} \frac{{\left(144\right)}^{{\displaystyle \frac{1}{2}}}+{\left(256\right)}^{{\displaystyle \frac{1}{2}}}}{3^2-2}= \\ \left(\mathrm{a}\right) 8 \left(\mathrm{b}\right) 4 \left(\mathrm{c}\right) -4 \left(\mathrm{d}\right) -8 \end{gathered}$$

Answer 11:

Since,

$$\begin{gathered} \frac{{\left(144\right)}^{{\displaystyle \frac{1}{2}}}+{\left(256\right)}^{{\displaystyle \frac{1}{2}}}}{3^2-2} \\ =\frac{{\left({12}^2\right)}^{{\displaystyle \frac{1}{2}}}+{\left({16}^2\right)}^{{\displaystyle \frac{1}{2}}}}{9-2} \\ =\frac{{12}^{2\times {\displaystyle \frac{1}{2}}}+{16}^{{\displaystyle 2\times \frac{1}{2}}}}{7} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =\frac{12+16}{7} \\ =\frac{28}{7} \\ =4 \end{gathered}$$

Hence, the correct alternative is option (b).

Page-6.33

Question 12:

Mark the correct alternative in the following question:

$$\begin{gathered} {\left(1+3+5+7+9+11\right)}^{\frac{3}{2}}= \\ \left(\mathrm{a}\right) 36 \left(\mathrm{b}\right) 216 \left(\mathrm{c}\right) 256 \left(\mathrm{d}\right) \mathrm{None} \mathrm{of} \mathrm{these} \end{gathered}$$

Answer 12:

Since,

$$\begin{gathered} {\left(1+3+5+7+9+11\right)}^{\frac{3}{2}} \\ ={\left(36\right)}^{\frac{3}{2}} \\ ={\left(6^2\right)}^{\frac{3}{2}} \\ =6^{2\times \frac{3}{2}} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =6^3 \\ =216 \end{gathered}$$

Hence, the correct alternative is option (b).

Question 13:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} abc=0, \mathrm{then} \frac{{\left\{{\left(x^a\right)}^b\right\}}^c}{{\left\{{\left(x^b\right)}^c\right\}}^a}= \\ \left(\mathrm{a}\right) 3 \left(\mathrm{b}\right) 0 \left(\mathrm{c}\right) -1 \left(\mathrm{d}\right) 1 \end{gathered}$$

Answer 13:

$$\begin{gathered} \frac{{\left\{{\left(x^a\right)}^b\right\}}^c}{{\left\{{\left(x^b\right)}^c\right\}}^a} \\ =\frac{{\left(x^a\right)}^{bc}}{{\left(x^b\right)}^{ac}} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =\frac{x^{abc}}{x^{abc}} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =\frac{x^0}{x^0} \left[\mathrm{As}, abc=0\right] \\ =\frac{1}{1} \left[\mathrm{As}, x^0=1\right] \\ =1 \end{gathered}$$

Question 14:

Mark the correct alternative in the following question:

$$\begin{gathered} {\left(2^3\right)}^4= \\ \left(\mathrm{a}\right) 2^{4^3} \left(\mathrm{b}\right) 2^{3^4} \left(\mathrm{c}\right) {\left(2^4\right)}^3 \left(\mathrm{d}\right) \mathrm{None} \mathrm{of} \mathrm{these} \end{gathered}$$

Answer 14:

Since,

$$\begin{gathered} {\left(2^3\right)}^4 \\ =2^{3\times 4} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =2^{4\times 3} \\ ={\left(2^4\right)}^3 \left[\mathrm{As}, x^{mn}={\left(x^m\right)}^n\right] \end{gathered}$$

Hence, the correct alternative is option (c).

Question 15:

Mark the correct alternative in the following question:

$$\begin{gathered} {\left\{{\left(33\right)}^2-{\left(31\right)}^2\right\}}^{\frac{5}{7}}= \\ \left(\mathrm{a}\right) 64 \left(\mathrm{b}\right) 16 \left(\mathrm{c}\right) 32 \left(\mathrm{d}\right) 4 \end{gathered}$$

Answer 15:

Since,

$$\begin{gathered} {\left\{{\left(33\right)}^2-{\left(31\right)}^2\right\}}^{\frac{5}{7}} \\ ={\left\{1089-961\right\}}^{\frac{5}{7}} \\ ={\left\{128\right\}}^{\frac{5}{7}} \\ ={\left\{2^7\right\}}^{\frac{5}{7}} \\ =2^{7\times \frac{5}{7}} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =2^5 \\ =32 \end{gathered}$$

Hence, the correct alternative is option (c).

Question 16:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} abc=0, \mathrm{then} \mathrm{find} \mathrm{the} \mathrm{value} \mathrm{of} {\left\{{\left(x^a\right)}^b\right\}}^c. \\ \left(\mathrm{a}\right) 1 \left(\mathrm{b}\right) a \left(\mathrm{c}\right) b \left(\mathrm{d}\right) c \end{gathered}$$

Answer 16:

Since,

$$\begin{gathered} {\left\{{\left(x^a\right)}^b\right\}}^c \\ ={\left(x^a\right)}^{bc} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =x^{abc} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =x^0 \left[\mathrm{As}, abc=0\right] \\ =1 \left[\mathrm{As}, x^0=1\right] \end{gathered}$$

Hence, the correct alternative is option (a).

Question 17:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} a=3^{-3}-3^3 \mathrm{and} b=3^3-3^{-3}, \mathrm{then} \frac{a}{b}-\frac{b}{a}= \\ \left(\mathrm{a}\right) 0 \left(\mathrm{b}\right) 1 \left(\mathrm{c}\right) -1 \left(\mathrm{d}\right) 2 \end{gathered}$$

Answer 17:

$$\begin{gathered} \mathrm{Since}, a=3^{-3}-3^3 \\ =\frac{1}{3^3}-3^3 \left(\mathrm{As}, x^{-m}=\frac{1}{x^m}\right) \\ =\frac{1}{27}-\frac{27}{1} \\ =\frac{1}{27}-\frac{27\times 27}{1\times 27} \\ =\frac{1}{27}-\frac{729}{27} \\ =\frac{1-729}{27} \\ =\frac{-728}{27} \\ \mathrm{Also}, b=3^3-3^{-3} \\ =3^3-\frac{1}{3^3} \left(\mathrm{As}, x^{-m}=\frac{1}{x^m}\right) \\ =\frac{27}{1}-\frac{1}{27} \\ =\frac{27\times 27}{1\times 27}-\frac{1}{27} \\ =\frac{729}{27}-\frac{1}{27} \\ =\frac{729-1}{27} \\ =\frac{728}{27} \\ \mathrm{Now}, \\ \frac{a}{b}-\frac{b}{a} \\ =\frac{\left({\displaystyle \frac{-728}{27}}\right)}{\left({\displaystyle \frac{728}{27}}\right)}-\frac{\left({\displaystyle \frac{728}{27}}\right)}{\left({\displaystyle \frac{-728}{27}}\right)} \\ =\left(-1\right)-\left(-1\right) \\ =-1+1 \\ =0 \end{gathered}$$

Hence, the correct alternative is option (a).

Question 18:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{What} \mathrm{should} \mathrm{be} \mathrm{multiplied} \mathrm{to} 6^{-2} \mathrm{so} \mathrm{that} \mathrm{the} \mathrm{product} \mathrm{may} \mathrm{be} \mathrm{equal} \mathrm{to} 216? \\ \left(\mathrm{a}\right) 6^4 \left(\mathrm{b}\right) 6^5 \left(\mathrm{c}\right) 6^3 \left(\mathrm{d}\right) 6 \end{gathered}$$

Answer 18:

$$\begin{gathered} \mathrm{Let} 6^{\mathit{m}} \mathrm{should} \mathrm{be} \mathrm{multiplied} \mathrm{to} 6^{-2}. \\ \mathrm{A}.\mathrm{T}.\mathrm{Q}. \\ {6}^{-2}\times 6^m=216 \\ \Rightarrow 6^{-2+m}=6^3 \\ \mathrm{Comparing} \mathrm{th} \mathrm{exponent} \mathrm{of} \mathrm{both} \mathrm{the} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ -2+m=3 \\ \Rightarrow m=3+2 \\ \therefore m=5 \end{gathered}$$

So, 6⁵ should be multiplied.

Hence, the correct alternative is option (b).

Question 19:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} xyz=0, \mathrm{then} \mathrm{find} \mathrm{the} \mathrm{value} \mathrm{of} {\left(a^x\right)}^{yz}+{\left(a^y\right)}^{zx}+{\left(a^z\right)}^{xy}= \\ \left(\mathrm{a}\right) 3 \left(\mathrm{b}\right) 2 \left(\mathrm{c}\right) 1 \left(\mathrm{d}\right) 0 \end{gathered}$$

Answer 19:

Since,

$$\begin{gathered} {\left(a^x\right)}^{yz}+{\left(a^y\right)}^{zx}+{\left(a^z\right)}^{xy} \\ =a^{xyz}+a^{yzx}+a^{zxy} \left[\mathrm{As}, {\left(x^m\right)}^n=x^{mn}\right] \\ =a^{xyz}+a^{xyz}+a^{xyz} \\ =a^0+a^0+a^0 \\ =1+1+1 \left(\mathrm{As}, x^0=1\right) \\ =3 \end{gathered}$$

Hence, the correct option is (a).

Question 20:

Choose the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} 2^n=4096, \mathrm{then} 2^{n-5}= \\ \left(\mathrm{a}\right) 128 \left(\mathrm{b}\right) 64 \left(\mathrm{c}\right) 256 \left(\mathrm{d}\right) 32 \end{gathered}$$

Answer 20:

$$\begin{gathered} \mathrm{As}, 2^n=4096 \\ \Rightarrow 2^{\mathit{n}}=2^{12} \\ \mathrm{Comparing} \mathrm{the} \mathrm{exponent} \mathrm{of} \mathrm{both} \mathrm{the} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ n=12 \\ \mathrm{Now}, 2^{n-5}=2^{12-5} \\ =2^7 \\ =128 \end{gathered}$$

Hence, the correct alternative is option (a).

Question 21:

Choose the correct alternative in the following question:

The number 4,70,394 in standard form is written as

(a) 4.70394 $\times$ 10⁵                   (b) 4.70394 $\times$ 10⁴                   (c) 47.0394 $\times$ 10⁴                   (d) 4703.94 $\times$ 10²

Answer 21:

Since, 4,70,394 = 4.70394 $\times$ 100000 = 4.70394 $\times$ 10⁵.

So, the number 4,70,394 in standard form is written as 4.70394 $\times$ 10⁵.

Hence, the correct alternative is option (a).

Question 22:

Choose the correct alternative in the following question:

The number 2.35 $\times$ 10⁴ in the usual form is written as

(a) 2.35 $\times$ 10³                            (b) 23500                            (c) 2350000                            (d) 235 $\times$ 10⁴

Answer 22:

Since, 2.35 $\times$ 10⁴ =  2.35 $\times$ 10000 = 23500

So, the number 2.35 $\times$ 10⁴ in the usual form is written as 23500.

Hence, the correct alternative is option (b).

Question 23:

Choose the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} 3^x=6561, \mathrm{then} 3^{x-3}= \\ \left(\mathrm{a}\right) 81 \left(\mathrm{b}\right) 243 \left(\mathrm{c}\right) 729 \left(\mathrm{d}\right) 27 \end{gathered}$$

Answer 23:

$$\begin{gathered} \mathrm{As}, 3^x=6561 \\ \Rightarrow 3^x=3^8 \\ \mathrm{Comparing} \mathrm{the} \mathrm{exponent} \mathrm{of} \mathrm{both} \mathrm{the} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ x=8 \\ \mathrm{Now}, 3^{x-3}=3^{8-3}=3^5=243 \end{gathered}$$

Hence, the correct alternative is option (b).

Question 24:

Choose the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} 2^n=1024, \mathrm{then} 2^{\left(\frac{n}{2}+2\right)}= \\ \left(\mathrm{a}\right) 64 \left(\mathrm{b}\right) 128 \left(\mathrm{c}\right) 256 \left(\mathrm{d}\right) 512 \end{gathered}$$

Answer 24:

$$\begin{gathered} \mathrm{As}, 2^n=1024 \\ \Rightarrow 2^n=2^{10} \\ \mathrm{Comparing} \mathrm{the} \mathrm{exponent} \mathrm{of} \mathrm{both} \mathrm{the} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ n=10 \\ \mathrm{Now}, \\ {2}^{\left(\frac{n}{2}+2\right)} \\ =2^{\left(\frac{10}{2}+2\right)} \\ =2^{\left(5+2\right)} \\ =2^7 \\ =128 \end{gathered}$$

Hence, the correct alternative is option (b).

Question 25:

Choose the correct alternative in the following question:

$$\begin{gathered} {\left(8^4+8^2\right)}^{\frac{1}{2}}= \\ \left(\mathrm{a}\right) 84 \left(\mathrm{b}\right) 8\sqrt{77} \left(\mathrm{c}\right) 72 \left(\mathrm{d}\right) 8\sqrt{65} \end{gathered}$$

Answer 25:

Since,

$$\begin{gathered} {\left(8^4+8^2\right)}^{\frac{1}{2}} \\ ={\left(8^{2+2}+8^2\right)}^{\frac{1}{2}} \\ ={\left(8^2\times 8^2+8^2\right)}^{\frac{1}{2}} \left(\mathrm{As}, x^{m+n}=x^m\times x^n\right) \\ ={\left[8^2\times \left(8^2+1\right)\right]}^{\frac{1}{2}} \left[\mathrm{As}, ab+ac=a\times \left(b+c\right)\right] \\ ={\left(8^2\right)}^{\frac{1}{2}}\times {\left(8^2+1\right)}^{\frac{1}{2}} \left[\mathrm{As}, {\left(ab\right)}^m=a^m\times b^m\right] \\ =8^{2\times \frac{1}{2}}\times {\left(64+1\right)}^{\frac{1}{2}} \\ =8\times {65}^{\frac{1}{2}} \\ =8\sqrt{65} \end{gathered}$$

Hence, the correct alternative is option (d).