Question

Pls solve this
Not getting this one ​

Answer 1

Solution :-

$\huge \sf \bigg[ \bigg(2^{ \sqrt{x} + 3 } \bigg)^{ \frac{1}{2 \sqrt{x} } } \bigg]^{\frac{2}{ \sqrt{x} - 1} } = 2 \\ \\ \\ \huge \sf \bigg(2^{ \sqrt{x} + 3 } \bigg)^{ \frac{1}{2 \sqrt{x}} \times \frac{2}{ \sqrt{x} - 1 } } = 2 \\ \\ \\ \bf \because ({a}^{m} )^n = {a}^{mn} \\ \\ \\ \huge \sf \bigg(2^{ \sqrt{x} + 3 } \bigg)^{ \frac{2}{ 2 \sqrt{x } (\sqrt{x} - 1) } } = 2 \\ \\ \\ \huge \sf \bigg(2^{ \sqrt{x} + 3 } \bigg)^{ \frac{2}{ 2x - 2 \sqrt{x}}} = 2 \\ \\ \\ \huge \sf 2^{ \sqrt{x} + 3 \times \frac{2}{ 2x - 2 \sqrt{x}}} = 2 \\ \\ \\ \huge \sf 2^{\frac{2 \sqrt{x} + 6 }{ 2x - 2 \sqrt{x}}} = 2^1$

$\sf \dfrac{2 \sqrt{x} + 6 }{2x - 2 \sqrt{x} } =1 \\ \\ \\ \bf \because {a}^{m} = {a}^{n} \implies m = n \\ \\ \\ \sf 2 \sqrt{x} + 6 = 2x - 2 \sqrt{x} \\ \\ \\ \sf 2 \sqrt{x} + 2 \sqrt{x} = 2x - 6 \\ \\ \\ \sf 4 \sqrt{x} = 2(x - 3) \\ \\ \\ \sf 2 \sqrt{x} = x - 3$

Squaring on both sides

⇒ (2√x)² = (x - 3)²

⇒ 4x = x² + 3² - 2(x)(3)

Since (a - b)² = a² + b² - 2ab

⇒ 4x = x² + 9 - 6x

⇒ 0 = x² + 9 - 6x - 4x

⇒ 0 = x² + 9 - 10x

⇒ 0 = x² - 10x + 9

⇒ x² - 10x + 9 = 0

Splitting the middle term

⇒ x² - 9x - x + 9 = 0

⇒ x(x - 9) - 1(x - 9) = 0

⇒ (x - 1)(x - 9) = 0

⇒ x - 1 = 0 or x - 9 = 0

⇒ x = 1 or x = 9

Value of x cannot be 1 because the power of it becomes 1/0 and in which the LHS is not equal to 2.

So value of x is 9.