Given:\( \pi \) is defined as the ratio of the circumference (say \( c \) ) of a circle to its diameter (say \( d \) ). That is, \( \pi=\frac{c}{d} \).\( \pi \) is irrational.To do:We have to resolve the above contradiction.Solution:When we measure the values of $c$ and ... Read More

Given: Given number is $\sqrt{9.3}$.To do: We have to represent $\sqrt{9.3}$ on the number line.Solution:1. Draw a line segment $AB=9.3$ units.2. Produce $B$ till point $C$, such that $BC=1$ unit.3. Find the mid-point of $\mathrm{AC}$, let it be $\mathrm{O}$.4. Taking $O$ as the centre, draw a semi-circle, passing through $A$ and $C$.5. ... Read More

To do:We have to find the values of (i) \( 64^{\frac{1}{2}} \)(ii) \( 32^{\frac{1}{5}} \)(iii) \( 125^{\frac{1}{3}} \)Solution: We know that, $(a^m)^n=(a)^{mn}$Therefore, (i) $64^{\frac{1}{2}}=(8\times8)^{\frac{1}{2}}$$=(8^2)^{\frac{1}{2}}$$=(8)^{2\times\frac{1}{2}}$$=8^1$$=8$Hence $64^{\frac{1}{2}}=8$(ii) $32^{\frac{1}{5}}=(2\times2\times2\times2\times2)^{\frac{1}{5}}$$=(2^5)^{\frac{1}{5}}$$=(2)^{5\times\frac{1}{5}}$$=2^1$$=2$Hence $32^{\frac{1}{5}}=2$(iii) $125^{\frac{1}{3}}=(5\times5\times5)^{\frac{1}{3}}$$=(5^3)^{\frac{1}{3}}$$=(5)^{3\times\frac{1}{3}}$$=5^1$$=5$Hence $125^{\frac{1}{3}}=5$Read More

To do:We have to find the values of (i) \( 9^{\frac{3}{2}} \)(ii) \( 32^{\frac{2}{5}} \)(iii) \( 16^{\frac{3}{4}} \)(iv) \( 125^{\frac{-1}{3}} \)Solution: We know that, $(a^m)^n=(a)^{mn}$Therefore, (i) $9^{\frac{3}{2}}=(3\times3)^{\frac{3}{2}}$$=(3^2)^{\frac{3}{2}}$$=(3)^{2\times\frac{3}{2}}$$=3^3$$=27$Hence $9^{\frac{3}{2}}=27$(ii) $32^{\frac{2}{5}}=(2\times2\times2\times2\times2)^{\frac{2}{5}}$$=(2^5)^{\frac{2}{5}}$$=(2)^{5\times\frac{2}{5}}$$=2^2$$=4$Hence $32^{\frac{2}{5}}=4$(iii) $16^{\frac{3}{4}}=(2\times2\times2\times2)^{\frac{3}{4}}$$=(2^4)^{\frac{3}{4}}$$=(2)^{4\times\frac{3}{4}}$$=2^3$$=8$Hence $16^{\frac{3}{4}}=8$(iv) $125^{\frac{-1}{3}}=(5\times5\times5)^{\frac{-1}{3}}$$=(5^3)^{\frac{-1}{3}}$$=(5)^{3\times\frac{-1}{3}}$$=5^{-1}$$=\frac{1}{5}$Hence $125^{\frac{-1}{3}}=\frac{1}{5}$Read More

To do:We have to simplify(i) \( 2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} \)(ii) \( \left(\frac{1}{3^{3}}\right)^{7} \)(iii) \( \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} \)(iv) \( 7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} \)Solution: We know that, $(a^m)^n=(a)^{mn}$$a^m \times a^n=a^{m+n}$Therefore, (i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=(2)^{\frac{2}{3}+\frac{1}{5}}$$=(2)^{\frac{2\times5+1\times3}{15}}$              (LCM of 3 and 5 is 15)$=(2)^{\frac{10+3}{15}}$$=(2)^{\frac{13}{15}}$Hence $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=(2)^{\frac{13}{15}}$(ii) $(\frac{1}{3^{3}})^{7}=(3^{-3})^{7}$        ... Read More

Given: $(i)$. $3^2\times3^4\times3^8$$(ii)$. $6^{15}\div6^{10}$$(iii)$. $a^3\times a^2$$(iv)$. $7^x\times7^2$$(v)$. $(5^2)^3\div5^3$$(vi)$. $2^5\times5^5$$(vii)$. $a^4\times b^4$$(viii)$. $(3^4)^3$$(ix)$. $(2^{20}\div2^{15})\times2^3$$(x)$. $8^t\div8^2$To do: To simplify and write the answer in exponential form by using laws of exponents.Solution:$(i)$. $3^2\times3^4\times3^8$$=[3^{2+4+8}]$            [As we know that $a^m\times a^n=a^{m+n}$]$=3^{14}$$(ii)$. $6^{15}\div6^{10}$$=[6^{15-10}]$                    $[a^m\div a^n=a^{m-n}]$$=6^5$$(iii)$. $a^3\times a^2$$=[a^{3+2}]$  ... Read More

Given: $(i)$. $\frac{2^3\times3^4\times4}{3\times32}$$(ii)$. $[(5^2)^{3\ }\times5^4]\div5^7$$(iii)$. $(25^4\div5^3)$$(iv)$. $\frac{3\times7^2\times11^8}{21\times11^3}$$(v)$. $\frac{3^7}{3^4\times3^3}$$(vi)$. $2^0+3^0+4^0$$(vii)$. $2^0\times3^0\times4^0$$(viii)$. $(3^0+2^0)\times5^0$$(ix)$. $\frac{2^8\times a^5}{4^3\times a^3}$$(x)$. $(\frac{a^5}{a^3})\times a^8$$(xi)$. $\frac{4^5\times a^8b^3}{4^5\times a^5b^2}$$(xii)$. $(2^3\times2)^2$To do: To simplify and express each of the above in exponential form.Solution:$(i)$. $\frac{2^3\times3^4\times4}{3\times32}$$=\frac{2^3\times3^4\times2^2}{3\times2^5}$$=\frac{2^{3+2}\times3^4}{3\times2^5}$       $[\because\ a^m\times a^n=a^{m+n}]$$=\frac{2^5\times3^4}{3\times2^5}$$=2^{5-5}\times 3^{4-1}$$=2^0\times3^3$                              ... Read More

Given: $(i)$. $10\times10^{11}=100^{11}$$(ii)$. $2^3$>$5^2$$(iii)$. $2^3\times3^2=6^5$$(iv)$. $3^0=(1000)^0$ To do: To state true or false and justify the answer. Solution:  $(i)$. $10\times10^{11}=100^{11}$ LHS $=(10^{1+11})$                                          $a^m\times a^n=a^{m+n}$ $=10^{12}$RHS $==(10^2)^11$$= (10)^{2\times11}$$= 10^{22}$ Thus the above equation is false. $(ii)$. $2^3>5^2$ Or $2\times2\times2$>$5\times5$ Or $8>25$ But 8

Given: $(i)$. $108\times192$$(ii)$. $270$$(iii)$. $729\times64$$(iv)$. $768$To do: To express each of the above as the product of prime factors only in exponential form.Solution: $(i)$. $108\times192$$=(2\times 2\times 3\times 3\times 3)\times(2\times2\times2\times2\times2\times2\times3)$$=(2^2\times3^3)\times(2^6\times3)$$=2^{2+6}\times3^{3+1}$$=2^8\times3^4$$(ii)$. $270$$=2\times3\times3\times3\times5$$=2\times3^3\times5$$(iii)$. $729\times64$$=3\times3\times3\times3\times3\times3\times2\times2\times2\times2\times2\times2$$=3^6\times2^6$$(iv)$. $768$$=2\times2\times2\times2\times2\times2\times2\times2\times3$$=2^8\times3$Read More

Given: $(i)$. $\frac{(2^5)^2\times7^3}{8^3\times7}$$(ii)$. $\frac{25\times5^2\times t^8}{10^3\times t^4}$$(iii)$. $\frac{3^5\times10^5\times25}{5^7\times6^5}$To do: To simplify each of the above expressions.Solution: $(i)$. $\frac{(2^5)^2\times7^3}{8^3\times7}$$=\frac{(2^{5})^2\times7^3}{(2^3)^3\times7}$$=\frac{2^{10}\times7^3}{2^9\times7}$$=(2^{10-9})\times(7^{3-1})$$=2^1\times7^2$$=2\times49$$=98$$(ii)$. $\frac{25\times5^2\times t^8}{10^3\times t^4}$$=\frac{5^2\times5^2\times t^8}{(5\times2)^3\times t^4}$$=\frac{5^{2+2}\times t^{8-4}}{5^3\times2^3}$$=\frac{5^{4-3}\times t^4}{2^3}$$=\frac{5^1\times t^4}{8}$$=\frac{5t^4}{8}$$(iii)$. $\frac{3^5\times10^5\times25}{5^7\times6^5}$$=\frac{3^5(2\times5)^5\times5^2}{5^7\times(2\times3)^5}$$=\frac{(3^5\times2^5\times5^{5+2})}{5^7\times2^5\times3^5}$$=\frac{3^5\times2^5\times5^7}{5^7\times2^5\times3^5}$$=2^0\times3^0\times5^0$$=1\times1\times1$$=1$Read More