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GenaCL600 [577]
2 years ago
11

%5E%7Bx%7D%20%7D%20%20-%20%20%20%5Csqrt%7B%20%7B9%7D%5E%7Bx%20%7D%20%7D%20" id="TexFormula1" title=" {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {9}^{x } } " alt=" {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {9}^{x } } " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
strojnjashka [21]2 years ago
5 0

Answer:

x = 0

Step-by-step 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 \times 3)}^{x}  \\  \\   {2}^{2x}  =  {2 ^{x}  \times 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(  \frac{2}{3} \bigg)^{x}  = 1 \\  \\  \implies \: x = 0 \\  \\  \because \: for \: x = 0 \\  \\ \bigg(  \frac{2}{3} \bigg)^{0}  = 1

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