What is the answer to 2^x+2^{-x}=7 ?

The answer to 2^x+2^{-x}=7 is x=(ln((7+3sqrt(5))/2))/(ln(2)),x=(ln((7-3sqrt(5))/2))/(ln(2))

Solve for x: 2^x+2^{-x}=7

The solution is x=(ln((7+3sqrt(5))/2))/(ln(2)),x=(ln((7-3sqrt(5))/2))/(ln(2))

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2^{x} + 2^{- x} = 7

x = \frac{ln ( \frac{7 + 3 5}{2} )}{ln ( 2 )}, x = \frac{ln ( \frac{7 - 3 5}{2} )}{ln ( 2 )}

Solution steps

2^{x} + 2^{- x} = 7

Apply exponent rules

2^{x} + (2^{x})^{- 1} = 7

Rewrite the equation with

2^{x} = u

u + (u)^{- 1} = 7

Solve

u + u^{- 1} = 7 : u = \frac{7 + 3 5}{2}, u = \frac{7 - 3 5}{2}

u = \frac{7 + 3 5}{2}, u = \frac{7 - 3 5}{2}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = \frac{7 + 3 5}{2} : x = \frac{ln ( \frac{7 + 3 5}{2} )}{ln ( 2 )}

Solve

2^{x} = \frac{7 - 3 5}{2} : x = \frac{ln ( \frac{7 - 3 5}{2} )}{ln ( 2 )}

x = \frac{ln ( \frac{7 + 3 5}{2} )}{ln ( 2 )}, x = \frac{ln ( \frac{7 - 3 5}{2} )}{ln ( 2 )}

5^{x} = 9

x, e^{2 x} - 3 e^{x} + 2 = 0

x - 2^{- x} = 0

x, e^{x y} + y = x - 1

2 e^{2 x} - e^{- x} = 0

What is the answer to 2^x+2^{-x}=7 ?
The answer to 2^x+2^{-x}=7 is x=(ln((7+3sqrt(5))/2))/(ln(2)),x=(ln((7-3sqrt(5))/2))/(ln(2))
Solve for x: 2^x+2^{-x}=7
The solution is x=(ln((7+3sqrt(5))/2))/(ln(2)),x=(ln((7-3sqrt(5))/2))/(ln(2))