Q: Simplify: \$\displaystyle \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\$

\$ \cfrac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\$\$=\cfrac { 3^{ -5 }\times (2\times 5)^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times (2\times 3)^{ -5 } } \$\$=\cfrac { 3^{ -5 }\times 2^{ -5 }\times 5^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times 2^{ -5 }\times 3^{ -5 } } \$\$=3^{ -5+5 }\times 2^{ -5+5 }\times 5^{ -5+3+7 }\$\$=5^{5}\$\$=3125\$

Q: \${5^{x - 3}}\,\,\, \times \,{3^{2x - 8}} = 225\$

We have, \${{5}^{x-3}}\times {{3}^{2x-8}}=225\$ \$ \dfrac{{{5}^{x}}}{{{5}^{3}}}\times \dfrac{{{3}^{2x}}}{{{3}^{8}}}=225 \$ \$ {{5}^{x}}\times {{3}^{2x}}=25\times 9\times {{5}^{3}}\times {{3}^{8}} \$ \$ {{5}^{x}}\times {{3}^{2x}}={{5}^{2}}\times {{3}^{2}}\times {{5}^{3}}\times {{3}^{8}} \$ \$ {{5}^{x}}\times {{3}^{2x}}={{5}^{5}}\times {{3}^{10}} \$ \$ {{5}^{x}}\times {{3}^{2x}}={{5}^{5}}\times {{3}^{2\times 5}} \$ So, \$x = 5\$ Hence, the value of \$x\$ is \$5\$.

Q: Find the value of \$m\$ for which \$5^m \div 5^{-3} = 5^5\$

\$5^m \div 5^{-3} = 5^5\$\$\Rightarrow \displaystyle 5^m \div \frac{1}{5^3} = 5^5\$\$ \Rightarrow 5^m \times 5^3 = 5^5\$\$\Rightarrow 5^{m + 3} = 5^5\$\$\Rightarrow m + 3 = 5\$\$\Rightarrow m = 5 - 3 = 2\$

Q: If \$a^x=b,b^y=c,c^z=a\$ then show that \$xyz=1\$.

\$a^x=b\$......(1), \$b^y=c\$......(2), \$c^z=a\$......(3).Now putting the value of \$c\$ from (2) in (3) we get,\$b^{yz}=a\$Now using the value of \$b\$ from (1) in the above equation we get,\$a^{xyz}=a\$or, \$xyz=1\$.

Q: Simplify \$\dfrac { 4 }{ { \left( 216 \right) }^{ -2/3 } } +\dfrac { 1 }{ { \left( 256 \right) }^{ -3/4 } } +\dfrac { 2 }{ { \left( 243 \right) }^{ -1/5 } }\$

\$\dfrac{4}{{{{\left( {216} \right)}^{ - 2/3}}}} + \dfrac{1}{{{{\left( {216} \right)}^{ - 3/4}}}} + \frac{2}{{{{\left( {216} \right)}^{ - 1/5}}}}\$\$ = \dfrac{4}{{{{\left( {{6^3}} \right)}^{ - 2/3}}}} + \dfrac{1}{{{{\left( {{4^4}} \right)}^{ - 3/4}}}} + \dfrac{2}{{{{\left( {{3^5}} \right)}^{ - 1/5}}}}\$\$ = \dfrac{4}{{{6^{ - 2}}}} + \dfrac{1}{{{4^{ - 3}}}} + \dfrac{2}{{{3^{ - 1}}}}\$\$ = 4 \times {6^2} + {4^3} + 2 \times 3\$\$ = 4 \times 36 + 64 + 6\$\$ = 144 + 70\$\$ = 214\$

Q: Simplify and express the result in power notation with positive exponent.(i) \$(-4)^{5}\div (-4)^{8} \$(ii) \$\left(\dfrac{1}{2^{3}}\right)^{2} \$(iii) \$(-3)^{4}\times\left(\dfrac{5}{3}\right)^{4} \$(iv) \$\left(3^{-7}\div 3^{-10}\right)\times 3^{-5} \$(v) \$2^{-3}\times(-7)^{(-3)} \$

(i) \$(-4)^{5}\div (-4)^{8} = \dfrac{(-4)^5}{(-4)^8} = (-4)^{5-8} = (-4)^{-3} =\dfrac {1}{(-4)^3}\$(ii) \$\left (\dfrac{1}{2^{3}}\right )^{2} = \dfrac{1}{2^{3\times2}} =\dfrac{1}{2^{6}} \$(iii) \$(-3)^{4}\times \left (\dfrac{5}{3}\right )^{4} = (-1)^4\times(3)^4\times\left(\dfrac53\right)^4 = (-1)^{4}\times 5^{4} =5^{4} \$(iv) \$(3^{-7}\div 3^{-10})\times 3^{-5} = \left(\dfrac{3^{-7}}{3^{-10}}\right)\times(3)^{-5} = 3^{-7+10}\times 3^{-5} \$ \$ = 3^{3-5} \$\$ = 3^{-2} = \dfrac{1}{3^{2}} \$(v) \$2^{-3}\times(-7)^{(-3)} = \dfrac1{2^3}\times\dfrac1{(-7)^{3}} = \left (\dfrac{1}{2\times (-7)}\right )^{3} \$ \$ =\dfrac{1}{(-14)^{3}} \$

Q: If \$\dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}\$Show that: \$m-n=1\$

\$\dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}\$\$\Rightarrow \dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}\$\$\Rightarrow \dfrac{3^{3n}.3^2-3^{3n}}{3^{3m}.2^3}=\dfrac{1}{3^3}\$\$\Rightarrow \dfrac{3^{3n}(3^2-1)}{3^3m\times 8}=\dfrac{1}{3^3}\$\$\Rightarrow \dfrac{3^{3n}\times 8}{3^3m\times 8}=\dfrac{1}{3^3}\$\$\Rightarrow \dfrac{1}{3^{3(m-n)}}=\dfrac{1}{3^{3-1}}\$\$\Rightarrow m-n=1\$ (proved)

Q: Evaluate : \$4\sqrt { \left( 81 \right) ^{ -2 } } \$

\$\sqrt[4]{(81)^{-2}}=81^{2\times \dfrac{1}{4}}\$\$=81^{-\dfrac{1}{2}}=(\sqrt{81})^{-1}=9^{-1}\$\$=\dfrac{1}{9}\$.

Q: By what number should \$\displaystyle \left ( \frac{-2}{3} \right )^{-3}\$ be divided so that the quotient may be \$\displaystyle \left ( \frac{4}{27} \right )^{-2}\$

Let the number be \$x\$ \$\displaystyle \frac{ \left ( \frac{-2}{3} \right )^{-3}}{x}= \left ( \frac{4}{27} \right )^{-2}\$\$\displaystyle\Rightarrow\frac{ \left ( \frac{-3}{2} \right )^{3}}{x}= \left ( \frac{27}{4} \right )^{2}\$\$\displaystyle\Rightarrow x=\dfrac{(\frac{-3}{2} )^{3}}{(\frac{27}{4} )^{2}}\$\$\displaystyle\Rightarrow x=\dfrac{(\frac{-27}{8} )}{(\frac{27^2}{16} )}\$\$\displaystyle\Rightarrow x=\dfrac{-2}{27}\$

Q: \$a^{m}\times a^{n}\$ is equal to

Reffer image.So, \$a^{m}\$ x \$a^{n}=a^{(m+n)}\$

Question 1

Simplify: $\displaystyle \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$

Accepted Answer

$ \cfrac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$$=\cfrac { 3^{ -5 }\times (2\times 5)^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times (2\times 3)^{ -5 } } $$=\cfrac { 3^{ -5 }\times 2^{ -5 }\times 5^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times 2^{ -5 }\times 3^{ -5 } } $$=3^{ -5+5 }\times 2^{ -5+5 }\times 5^{ -5+3+7 }$$=5^{5}$$=3125$

Question 2

${5^{x - 3}}\,\,\, \times \,{3^{2x - 8}} = 225$

Accepted Answer

We have,&#10;&#10;${{5}^{x-3}}\times&#10;{{3}^{2x-8}}=225$&#10;&#10;$ \dfrac{{{5}^{x}}}{{{5}^{3}}}\times&#10;\dfrac{{{3}^{2x}}}{{{3}^{8}}}=225 $&#10;&#10;$ {{5}^{x}}\times&#10;{{3}^{2x}}=25\times 9\times {{5}^{3}}\times {{3}^{8}} $&#10;&#10;$ {{5}^{x}}\times&#10;{{3}^{2x}}={{5}^{2}}\times {{3}^{2}}\times {{5}^{3}}\times {{3}^{8}} $&#10;&#10;$ {{5}^{x}}\times&#10;{{3}^{2x}}={{5}^{5}}\times {{3}^{10}} $&#10;&#10;$ {{5}^{x}}\times&#10;{{3}^{2x}}={{5}^{5}}\times {{3}^{2\times 5}} $&#10;&#10; &#10;&#10;So, $x = 5$&#10;&#10; &#10;&#10;Hence, the value of $x$ is $5$.

Question 3

Find the value of $m$ for which $5^m \div 5^{-3} = 5^5$

Accepted Answer

$5^m \div 5^{-3} = 5^5$$\Rightarrow \displaystyle 5^m \div \frac{1}{5^3} = 5^5$$ \Rightarrow 5^m \times 5^3 = 5^5$$\Rightarrow 5^{m + 3} = 5^5$$\Rightarrow m + 3 = 5$$\Rightarrow m = 5 - 3 = 2$

Question 4

If $a^x=b,b^y=c,c^z=a$ then show that $xyz=1$.

Accepted Answer

$a^x=b$......(1), $b^y=c$......(2), $c^z=a$......(3).Now putting the value of $c$ from (2) in (3) we get,$b^{yz}=a$Now using the value of $b$ from (1) in the above equation we get,$a^{xyz}=a$or, $xyz=1$.

Question 5

Simplify $\dfrac { 4 }{ { \left( 216 \right)  }^{ -2/3 } } +\dfrac { 1 }{ { \left( 256 \right)  }^{ -3/4 } } +\dfrac { 2 }{ { \left( 243 \right)  }^{ -1/5 } }$

Accepted Answer

$\dfrac{4}{{{{\left( {216} \right)}^{ - 2/3}}}} + \dfrac{1}{{{{\left( {216} \right)}^{ - 3/4}}}} + \frac{2}{{{{\left( {216} \right)}^{ - 1/5}}}}$$ = \dfrac{4}{{{{\left( {{6^3}} \right)}^{ - 2/3}}}} + \dfrac{1}{{{{\left( {{4^4}} \right)}^{ - 3/4}}}} + \dfrac{2}{{{{\left( {{3^5}} \right)}^{ - 1/5}}}}$$ = \dfrac{4}{{{6^{ - 2}}}} + \dfrac{1}{{{4^{ - 3}}}} + \dfrac{2}{{{3^{ - 1}}}}$$ = 4 \times {6^2} + {4^3} + 2 \times 3$$ = 4 \times 36 + 64 + 6$$ = 144 + 70$$ = 214$

Question 6

Simplify and express the result in power notation with positive exponent.
(i) $(- 4)^{5} \div (- 4)^{8}$

(ii) $(\frac{1}{2 ^{3}})^{2}$

(iii) $(- 3)^{4} \times (\frac{5}{3})^{4}$

(iv) $(3^{- 7} \div 3^{- 10}) \times 3^{- 5}$

(v) $2^{- 3} \times (- 7)^{(- 3)}$

Medium

View solution

>

Accepted Answer

(i)    $(-4)^{5}\div (-4)^{8} = \dfrac{(-4)^5}{(-4)^8} = (-4)^{5-8} = (-4)^{-3} =\dfrac {1}{(-4)^3}$(ii)   $\left (\dfrac{1}{2^{3}}\right )^{2} = \dfrac{1}{2^{3\times2}} =\dfrac{1}{2^{6}} $(iii)   $(-3)^{4}\times \left (\dfrac{5}{3}\right )^{4} = (-1)^4\times(3)^4\times\left(\dfrac53\right)^4 =  (-1)^{4}\times 5^{4} =5^{4} $(iv)   $(3^{-7}\div 3^{-10})\times 3^{-5} = \left(\dfrac{3^{-7}}{3^{-10}}\right)\times(3)^{-5} = 3^{-7+10}\times 3^{-5} $ $ = 3^{3-5} $$ = 3^{-2} = \dfrac{1}{3^{2}} $(v)    $2^{-3}\times(-7)^{(-3)} = \dfrac1{2^3}\times\dfrac1{(-7)^{3}} = \left (\dfrac{1}{2\times (-7)}\right )^{3} $ $ =\dfrac{1}{(-14)^{3}} $

Question 7

If $\dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}$Show that: $m-n=1$

Accepted Answer

$\dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}$$\Rightarrow \dfrac{9^n.3^2.3^n-(27)^n}{(3^m.2)^3}=3^{-3}$$\Rightarrow \dfrac{3^{3n}.3^2-3^{3n}}{3^{3m}.2^3}=\dfrac{1}{3^3}$$\Rightarrow \dfrac{3^{3n}(3^2-1)}{3^3m\times 8}=\dfrac{1}{3^3}$$\Rightarrow \dfrac{3^{3n}\times 8}{3^3m\times 8}=\dfrac{1}{3^3}$$\Rightarrow \dfrac{1}{3^{3(m-n)}}=\dfrac{1}{3^{3-1}}$$\Rightarrow m-n=1$ (proved)

Question 8

Evaluate : $4\sqrt { \left( 81 \right) ^{ -2 } } $

Accepted Answer

$\sqrt[4]{(81)^{-2}}=81^{2\times \dfrac{1}{4}}$$=81^{-\dfrac{1}{2}}=(\sqrt{81})^{-1}=9^{-1}$$=\dfrac{1}{9}$.

Question 9

By what number should $\displaystyle \left ( \frac{-2}{3} \right )^{-3}$ be divided so that the quotient may be $\displaystyle \left ( \frac{4}{27} \right )^{-2}$

Accepted Answer

Let the number be $x$   $\displaystyle \frac{ \left ( \frac{-2}{3} \right )^{-3}}{x}= \left ( \frac{4}{27} \right )^{-2}$$\displaystyle\Rightarrow\frac{ \left ( \frac{-3}{2} \right )^{3}}{x}= \left ( \frac{27}{4} \right )^{2}$$\displaystyle\Rightarrow x=\dfrac{(\frac{-3}{2}  )^{3}}{(\frac{27}{4} )^{2}}$$\displaystyle\Rightarrow x=\dfrac{(\frac{-27}{8}  )}{(\frac{27^2}{16} )}$$\displaystyle\Rightarrow x=\dfrac{-2}{27}$

Question 10

$a^{m}\times a^{n}$ is equal to

Accepted Answer

Reffer image.So, $a^{m}$ x $a^{n}=a^{(m+n)}$

Revise with Concepts

Extending the Laws of Exponents

Powers with the Same Base

Power of a Power and Powers with Equal Exponents

Laws of Exponents Applied to Negative Exponents

Quick Summary With Stories

Powers with the Same Base

Power of a Power and Powers with Equal Exponents

Laws of Exponents Applied to Negative Exponents

Meaning of Negative Exponents

Dividing with the same Exponents

Multiplying Numbers with the Same Exponents or Powers

Dividing Powers with the Same Base

Taking power of a power

Laws of exponents

Simplify: $\frac{3 ^{- 5} \times 1 0 ^{- 5} \times 1 2 5}{5 ^{- 7} \times 6 ^{- 5}}$

$5^{x - 3} \times 3^{2 x - 8} = 225$

Find the value of $m$ for which $5^{m} \div 5^{- 3} = 5^{5}$

If $a^{x} = b, b^{y} = c, c^{z} = a$ then show that $x y z = 1$ .

Simplify $\frac{4}{( 2 1 6 ) ^{- 2 / 3}} + \frac{1}{( 2 5 6 ) ^{- 3 / 4}} + \frac{2}{( 2 4 3 ) ^{- 1 / 5}}$

Simplify and express the result in power notation with positive exponent.
(i) $(- 4)^{5} \div (- 4)^{8}$

(ii) $(\frac{1}{2 ^{3}})^{2}$

(iii) $(- 3)^{4} \times (\frac{5}{3})^{4}$

(iv) $(3^{- 7} \div 3^{- 10}) \times 3^{- 5}$

(v) $2^{- 3} \times (- 7)^{(- 3)}$

If $\frac{9 ^{n} . 3 ^{2} . 3 ^{n} - ( 2 7 ) ^{n}}{( 3 ^{m} . 2 ) ^{3}} = 3^{- 3}$
Show that: $m - n = 1$

Evaluate : $4 (81)^{- 2}$

By what number should $(\frac{- 2}{3})^{- 3}$ be divided so that the quotient may be $(\frac{4}{2 7})^{- 2}$

$a^{m} \times a^{n}$ is equal to

Revise with Concepts

Extending the Laws of Exponents

Powers with the Same Base

Power of a Power and Powers with Equal Exponents

Laws of Exponents Applied to Negative Exponents

Quick Summary With Stories

Powers with the Same Base

Power of a Power and Powers with Equal Exponents

Laws of Exponents Applied to Negative Exponents

Meaning of Negative Exponents

Dividing with the same Exponents

Multiplying Numbers with the Same Exponents or Powers

Dividing Powers with the Same Base

Taking power of a power

Laws of exponents

Simplify: 5−7×6−53−5×10−5×125​

5x−3×32x−8=225

Find the value of m for which 5m÷5−3=55

If ax=b,by=c,cz=a then show that xyz=1.

Simplify (216)−2/34​+(256)−3/41​+(243)−1/52​

Simplify and express the result in power notation with positive exponent. (i) (−4)5÷(−4)8 (ii) (231​)2 (iii) (−3)4×(35​)4 (iv) (3−7÷3−10)×3−5 (v) 2−3×(−7)(−3)

If (3m.2)39n.32.3n−(27)n​=3−3 Show that: m−n=1

Evaluate : 4(81)−2​

By what number should (3−2​)−3 be divided so that the quotient may be (274​)−2

am×an is equal to

Simplify: $\frac{3 ^{- 5} \times 1 0 ^{- 5} \times 1 2 5}{5 ^{- 7} \times 6 ^{- 5}}$

$5^{x - 3} \times 3^{2 x - 8} = 225$

Find the value of $m$ for which $5^{m} \div 5^{- 3} = 5^{5}$

If $a^{x} = b, b^{y} = c, c^{z} = a$ then show that $x y z = 1$ .

Simplify $\frac{4}{( 2 1 6 ) ^{- 2 / 3}} + \frac{1}{( 2 5 6 ) ^{- 3 / 4}} + \frac{2}{( 2 4 3 ) ^{- 1 / 5}}$

Simplify and express the result in power notation with positive exponent.
(i) $(- 4)^{5} \div (- 4)^{8}$

(ii) $(\frac{1}{2 ^{3}})^{2}$

(iii) $(- 3)^{4} \times (\frac{5}{3})^{4}$

(iv) $(3^{- 7} \div 3^{- 10}) \times 3^{- 5}$

(v) $2^{- 3} \times (- 7)^{(- 3)}$

If $\frac{9 ^{n} . 3 ^{2} . 3 ^{n} - ( 2 7 ) ^{n}}{( 3 ^{m} . 2 ) ^{3}} = 3^{- 3}$
Show that: $m - n = 1$

Evaluate : $4 (81)^{- 2}$

By what number should $(\frac{- 2}{3})^{- 3}$ be divided so that the quotient may be $(\frac{4}{2 7})^{- 2}$

$a^{m} \times a^{n}$ is equal to