Find:\( 6^{x}+9^{x}=2^{2 x+1} \)

1.	Find:\( 6^{x}+9^{x}=2^{2 x+1} \)
Answer» 6^x + 9^x = 2^{2x + 1} ⇒ 3^2x + 3^x.2^x = 2^{2x + 1} ⇒ 3^2x + 2.3^x .\((\frac{2^x}2)+\frac{2^{2x}}4\) = 2^2x+1 + \(\frac{2^{2x}}4\) ⇒ (3^x + \(\frac{2^x}2\))² = 2^2x\((2+\frac14)\) (\(\because\) a² + 2ab + b² = (a + b)²) ⇒ (3^x + 2^x-1) = 3/2 . 2^x = 3.2^x-1 ⇒ 3^x = 3.2^x-1- 2^x-1 = 2.2^x-1 ⇒ 3^x = 2^x which is possible only when x = 0 hence, if 6^x + 9^x = 2^2x+1 then x = 0

Answer»

6^x + 9^x = 2^{2x + 1}

⇒ 3^2x + 3^x.2^x = 2^{2x + 1}

⇒ 3^2x + 2.3^x .\((\frac{2^x}2)+\frac{2^{2x}}4\) = 2^2x+1 + \(\frac{2^{2x}}4\)

⇒ (3^x + \(\frac{2^x}2\))² = 2^2x\((2+\frac14)\) (\(\because\) a² + 2ab + b² = (a + b)²)

⇒ (3^x + 2^x-1) = 3/2 . 2^x = 3.2^x-1

⇒ 3^x = 3.2^x-1- 2^x-1 = 2.2^x-1

⇒ 3^x = 2^x

which is possible only when x = 0

hence, if 6^x + 9^x = 2^2x+1 then x = 0

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Find:\( 6^{x}+9^{x}=2^{2 x+1} \)

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