question_answer

Find the value of n, where n is an integer and \[{{2}^{n-5}}\times {{6}^{2n-4}}=\frac{1}{{{12}^{4}}\times 2}\].

Answer:

Given,         \[{{2}^{n-5}}\times {{6}^{2n-4}}=\frac{1}{{{12}^{4}}\times 2}\] \[{{2}^{n-5}}\times {{\left( 3\times 2 \right)}^{2n-4}}=\frac{1}{{{\left( 2\times 2\times 3 \right)}^{4}}\times 2}\]   \[\left[ \because {{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}} \right]\] \[{{2}^{n-5+\left( 2n-4 \right)}}\times {{3}^{2n-4}}=\frac{1}{{{2}^{4+4+1}}\times {{3}^{4}}}\] \[\left[ \because {{a}^{m}}\times {{a}^{n}}={{a}^{m}}+n \right]\]           \[{{2}^{3n-9}}\times {{3}^{2n-4}}=\frac{1}{{{2}^{9}}\times {{3}^{4}}}\] \[{{2}^{3n-9}}\times {{3}^{2n-4}}={{2}^{-9}}\times {{3}^{-4}}\] \[\left[ \because \frac{1}{{{a}^{m}}}={{a}^{-m}} \right]\] If bases are equal, then we can equate their powers. \[{{2}^{3n-9}}={{2}^{-9}}\Rightarrow 3n-3=-9\] \[3n=-9+9\Rightarrow 3n=0\Rightarrow n=0\]