Question

`{2}^(x) - {3}^(x) = \sqrt{ {6}^(x) } - \sqrt{ {9}^(x ) }`
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Answer 1

x = 0

Explanation:

`{2}^(x) - {3}^(x) = \sqrt{ {6}^(x) } - \sqrt{ {9}^(x ) } \\ \\ {2}^(x) - {3}^(x) = \sqrt{ {6}^(x) } - \sqrt{ {( {3}^(2)) }^(x ) } \\ \\ {2}^(x) - {3}^(x) = \sqrt{ {6}^(x) } - \sqrt{ {({3)}^(2x)}}\\ \\ {2}^(x) - \cancel{{3}^(x)} = \sqrt{ {6}^(x) } - \cancel{ {3}^(x) } \\ \\ {2}^(x) = \sqrt{ {6}^(x) } \\ squaring \: both \: sides \\ \\ {( {2}^(x) )}^(2) = {(\sqrt{ {6}^(x) })}^(2) \\ \\ {2}^(2x) = {6}^(x) \\ \\ {2}^(2x) = {(2 * 3)}^(x) \\ \\ {2}^(2x) = {2 ^(x) * 3}^(x) \\ \\ \frac{ {2}^(2x) }{ {2}^(x) } = 3^(x) \\ \\ {2}^(2x - x) = {3}^(x) \\ \\ {2}^(x) = {3}^(x) \\ \\ \frac{ {2}^(x) }{ {3}^(x) } = 1 \\ \\ \bigg( (2)/(3) \bigg)^(x) = 1 \\ \\ \implies \: x = 0 \\ \\ \because \: for \: x = 0 \\ \\ \bigg( (2)/(3) \bigg)^(0) = 1`