Answer:
Since, \[{{\left( \frac{4}{9} \right)}^{4}}\times {{\left( \frac{4}{9} \right)}^{-7}}={{\left( \frac{4}{9} \right)}^{2x-1}}\] or \[{{\left( \frac{4}{9} \right)}^{4+(-7)}}={{\left( \frac{4}{9} \right)}^{2x-1}}\] or \[{{\left( \frac{4}{3} \right)}^{-3}}={{\left( \frac{4}{9} \right)}^{2x-1}}\] By comparing with powers then, \[-~~3=2x1\] \[2x=-3+1\] or \[~2x=2\] or \[x=-\frac{2}{2}=-1\] Hence,\[x=1\]
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