Question

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Answer 1

❄ Question :

Q2) Find the value of X so that :

$\large\rm{(1) 3^2 \times (-4)^2=(-12)^2x }$

$\large\rm{(2) \bigg(\dfrac{-3}{2}\bigg) ^6 \times \bigg(\dfrac{4}{9}\bigg)^3=\bigg(\dfrac{1}{2}\bigg)^3x}$

$\rm(3) \large\: \bigg( { \frac{4}{5} } \bigg)^{ - 2} \div \bigg( { \frac{ - 4}{5} } \bigg) = {(1)}^{3x}$

$\rm \large \: (4) \: \bigg( { \frac{9}{4} } \bigg)^{3} \times \bigg( { \frac{8}{9} } \bigg)^{3} = {2}^{6x}$

$\rm \large \: (5) \: \bigg( { \frac{15}{4} } \bigg)^{3} \div \bigg( { \frac{5}{4} } \bigg)^{3} = {3}^{x}$

To find :

• The value of x=?

❄ Solution :

$\bf \large \: (1) \: {3}^{2} \times {( - 4)}^{2} = {( - 12)}^{2x}$

$\\ \implies \large \sf \: {3}^{2} \times {( - 4)}^{ 2} = {( - 12)}^{2x}$

$\\ \implies \large \sf \: \: 9 \times 16 = {( - 12)}^{2x}$

$\implies \large \sf \: 144 = {( - 12)}^{2x}$

$\\ \implies \large \sf \: {12}^{2} = { ( - 12)}^{2x}$

$\\ \implies \underline{ \sf{subtracting \: both \: side \: by \: ( - ) \: sign}}$

$\\ \implies \large \sf \: { - 12}^{2} = - ( { - 12)}^{2x}$

$\\ \: \implies \bf{ \underline{lhs = rhs \: are \: equal \: so}}$

$\\ \implies \large \sf \: 2x = 2$

$\\ \: \implies \large \sf \: \cancel{2}x = \cancel{2} \: \: \: \: \: \: \: \: \: \therefore\boxed{x = 1}$

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$\large \bf \: (2) \bigg( { \frac{ - 3}{2} } \bigg)^{6} \times \bigg( { \frac{4}{9} } \bigg)^{3} = \bigg({ \frac{1}{2} } \bigg)^{3x}$

$\\ \implies \large \sf \: \bigg( { \frac{ - 3}{2} } \bigg)^{6} \times \bigg( { \frac{4}{9} } \bigg)^{3} = \: \bigg( { \frac{1}{2} } \bigg)^{3x}$

$\\ \implies \large \sf \bigg( \frac{729}{64} \bigg) \times \bigg( \frac{64}{729} \bigg) = \bigg( { \frac{1}{2} } \bigg)^{3x}$

$\\ \implies \large \sf \: \bigg( \cancel{\frac{729}{64} } \bigg) \times \bigg( \cancel \frac{64}{729} \bigg) = \bigg( { \frac{1}{2} } \bigg)^{3x}$

$\\ \implies \large \sf \: \: 1 = \bigg({ \frac{1}{2} } \bigg)^{3x}$

$\\ \: \implies \large \sf \: \bigg( { \frac{1}{2} } \bigg)^{0} = \bigg( { \frac{1}{2} } \bigg)^{3x}$

$\: \bf \large{ \underline{ since \: \: \: } \: \: \: \: \: \: {\:\boxed{ {a}^{0} = 1}}}$

$\\ \implies \large \sf \: \: 0 = 3 x$

$\\ \implies \large \sf{ \underline{since \:} \: \: \: \: \: \: \: \: \boxed{if \: {a}^{m} = {a}^{n} \: then \: \: m = n}}$

$\implies \large \sf \boxed{x = 0 \: }$

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$\\ \bf \large \:(3) \bigg( { \frac{4}{5} } \bigg)^{ - 2} \div \bigg( { \frac{ - 4}{5} } \bigg)^{ - 2} = {1}^{2x}$

$\\ \implies \large \sf \: \bigg( { \frac{4}{5} } \bigg)^{ - 2} \div \bigg( { \frac{ - 4}{5} } \bigg)^{ - 2} = {1}^{2x}$

$\\ \implies \large \sf \bigg( { \frac{ \frac{ 4}{5} }{ \frac{ - 4}{5} } } \bigg)^{ - 2 + 2} = {1}^{2x}$

$\\ \implies \large \sf \: { - 1}^{0} = {1}^{2x}$

$\\ \implies \large \sf \: 1 = {1}^{2x}$

$\\ \implies \large \sf \: 2x = 1$

$\implies \large \boxed{ \underline{ \: x = \frac{1}{2} }}$

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$\\ \bf \large \: (4) \bigg({ \frac{9}{4} } \bigg)^{3} \times \bigg( { \frac{8}{9} } \bigg)^{3} = {2}^{6x}$

$\\ \implies \large \sf \bigg( \frac{729}{64} \bigg) \times \bigg( \frac{512}{729} \bigg) = {2}^{6x}$

$\\ \implies \large \sf \bigg( \frac{ \cancel{729}}{64} \bigg) \times \bigg( \frac{512}{ \cancel{729}} \bigg) = {2}^{6x}$

$\\ \implies \large \sf \bigg( \cancel\frac{512}{64} 8 \bigg) = {2}^{6x}$

$\\ \implies \large \sf \bigg( {2}^{3} \bigg) = {2}^{6x}$

$\\ \implies \large\sf \: 3 = 6x$

$\\ \implies \large \sf \: \cancel3 = \cancel6x$

$\implies \large \boxed{ \underline{ \sf{x = \frac{1}{2} }}}$

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$\\ \large \bf \: (5) \: \bigg( { \frac{15}{4} } \bigg)^{3} \div \bigg( { \frac{5}{4} } \bigg) ^{3} = {3}^{x}$

$\\ \implies \large \sf \bigg( { \frac{15}{4} } \bigg)^{3} \div \bigg( { \frac{5}{4} } \bigg)^{3} = {3}^{x}$

$\\ \implies \large \sf \bigg( { \frac{ \frac{15}{4} }{ \frac{5}{4} } } \bigg)^{3 - 3} = {3}^{x}$

$\\ \implies \large \sf \bigg( { \frac{ \frac{ \cancel{15} \: 3}{ \cancel{4}} }{ \cancel{ \frac{5}{4} }} } \bigg)^{ 0} = {3}^{x}$

$\\ \implies \large \sf \: {3}^{0} = {3}^{x}$

$\\ \implies \large \sf \: \: x = 0$

$\implies \large \boxed { \underline{ \sf{ \: x =0}}}$

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$\\ \red\huge\bigstar\boxed{\sf\color{red}{Additional \: information }}$

Laws of Indices :

$\\ \large\bf {(1) a^m . a^n= a^{m+n}}$

$\\ \large\bf{ (2) (a^m)^n=a^{mn}}$

$\\ \large\bf{(3) (ab)^m=a^m.a^n}$