Question

4^1+x + 4^1-x = 10 then the value of "x" is

Answer 1

$HEYA \: \: \: \\ \\ 4 {}^{( 1 + x)} + 4 {}^{(1 - x)} = 10 \\ \\ 4 {}^{1} \times 4 {}^{x} + 4 {}^{1} \times 4 {}^{ - x} = 10 \\ \\ 4 {}^{x} + 4 {}^{ - x} = 10 \div 4 \\ \\ 4 {}^{x} + 4 {}^{ - x} = 5 \div 2 \\ \\ let \: \: 4 {}^{x} = z \: \: \\ \\ z + (1 \div z) = 5 \div 2 \\ \\ z { }^{2} + 1 = 5z \div 2 \\ \\ 2z {}^{2} - 5z + 2 = 0 \\ \\ 2z {}^{2} - 4z - z + 2 = 0 \\ \\ 2z(z - 2) - 1(z - 2) = 0 \\ \\ (2z - 1) = 0 \: \: or \: \: (z - 2) = 0 \\ \\ z = 1 \div 2 \: \: \: \: \: \: or \: \: \: \: \: \: \: z = 2 \\ \\ 4 {}^{x} = 2 {}^{ - 1} \: \: \: \: or \: \: \: \: 4 {}^{x} = 2 \\ \\ 2 {}^{2x} = 2 {}^{ - 1} \: \: \: \: or \: \: \: \: 2 {}^{2x} = 2 {}^{1} \\ now \: compare \: powers \: of \: \: 2 \: \: we \: have \: \\ \\ 2x = - 1 \: \: \: or \: \: \: 2x = 1 \\ \\ x = - 1 \div 2 \: \: \: \: or \: \: \: \: \: x = 1 \div 2$

Answer 2

Answer:

$x=$±$\frac{1}{2}$

Step-by-step explanation:

Given :

$4^{1+x}+4^{1-x} = 10$

find x,.

Solution :

$4^{1+x}+4^{1-x} = 10$

⇒ $4^{1} \times 4^{x}+4^{1} \times 4^{-x} = 10$

⇒ $4^{1} \times (4^{x}+4^{-x}) = 10$

⇒ $4(4^x+\frac{1}{4^x} ) = 10$

⇒ $\frac{4^{2x}+1}{4^x} = \frac{10}{4}$

⇒ $\frac{4^{2x}+1}{4^x} = \frac{5}{2}$

⇒ $2(4^{2x}+1) = 5(4^x)$

⇒ $2 \times 4^{2x} + 2 - 5 \times 4^x = 0$

⇒ $2 \times (4^{x})^2 - 5 \times 4^x + 2 = 0$

⇒ $2 \times (4^{x})^2 - 4 \times 4^x - 4^x + 2 = 0$

⇒ $2 \times 4^{x}[4^{x} - 2] - [4^x - 2] = 0$

⇒ $[2 \times 4^{x}-1][4^{x} - 2] = 0$

Either,

⇒ $[2 \times 4^{x}-1] = 0$

(or) (or both)

⇒ $[4^{x} - 2] = 0$

(i)

⇒ $[2 \times 4^{x}-1] = 0$

⇒ $2 \times 4^{x} = 1$

⇒ $4^x = \frac{1}{2}$

⇒ $4^x = \frac{1}{\sqrt{4}} = \frac{1}{4^{\frac{1}{2}}} = 4^{-\frac{1}{2}}$

⇒ $4^x = 4^{-\frac{1}{2}}$ ⇒ $x=-\frac{1}{2}$

(ii)

⇒ $[4^{x} - 2] = 0$

⇒ $4^{x} = 2$

⇒ $4^x = \sqrt{4}$

⇒ $4^{x} = 4^{\frac{1}{2}$

⇒ $x = \frac{1}{2}$