What is the answer to 2^{1+x}+2^{1-x}=5 ?

The answer to 2^{1+x}+2^{1-x}=5 is x=1,x=-1

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2^{1+x}+2^{1-x}=5

2^{1 + x} + 2^{1 - x} = 5

x = 1, x = - 1

Solution steps

2^{1 + x} + 2^{1 - x} = 5

Apply exponent rules

2^{x} \cdot 2^{1} + (2^{x})^{- 1} \cdot 2^{1} = 5

Rewrite the equation with

2^{x} = u

u \cdot 2^{1} + (u)^{- 1} \cdot 2^{1} = 5

Solve

u \cdot 2^{1} + u^{- 1} \cdot 2^{1} = 5 : u = 2, u = \frac{1}{2}

u = 2, u = \frac{1}{2}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 2 : x = 1

Solve

2^{x} = \frac{1}{2} : x = - 1

x = 1, x = - 1

x, \frac{( 9 ^{x} - 1 )}{( 3 ^{y} + 2 )} = 81

(\frac{1}{5})^{3 + x} = (\frac{1}{25})^{4}

1 1^{- x - 5} = 1 4^{3 x}

(\frac{1}{6})^{x} = (\frac{4}{3})^{x + 1}

x, 16 + 3^{2 x} = 3^{v} + 2

What is the answer to 2^{1+x}+2^{1-x}=5 ?
The answer to 2^{1+x}+2^{1-x}=5 is x=1,x=-1
Solve for x: 2^{1+x}+2^{1-x}=5
The solution is x=1,x=-1