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

The answer to 2^x+2^{-x}=5 is x=(ln((5+sqrt(21))/2))/(ln(2)),x=(ln((5-sqrt(21))/2))/(ln(2))

Solve for x: 2^x+2^{-x}=5

The solution is x=(ln((5+sqrt(21))/2))/(ln(2)),x=(ln((5-sqrt(21))/2))/(ln(2))

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

x = \frac{ln ( \frac{5 + 21}{2} )}{ln ( 2 )}, x = \frac{ln ( \frac{5 - 21}{2} )}{ln ( 2 )}

Decimal Notation

x = 2.26041 \dots, x = - 2.26041 \dots

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

2^{x} = u

u + (u)^{- 1} = 5

Solve

u + u^{- 1} = 5 : u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = \frac{5 + 21}{2} : x = \frac{ln ( \frac{5 + 21}{2} )}{ln ( 2 )}

Solve

2^{x} = \frac{5 - 21}{2} : x = \frac{ln ( \frac{5 - 21}{2} )}{ln ( 2 )}

x = \frac{ln ( \frac{5 + 21}{2} )}{ln ( 2 )}, x = \frac{ln ( \frac{5 - 21}{2} )}{ln ( 2 )}

2^{x} + 2^{- x} = 3

5^{n} \cdot (\frac{1}{5})^{3} = 125

4^{l o g_{2} (2 x)} = 36

k^{x + 1} \cdot k^{x - 1} = k^{10}

(3^{x}) + (3^{x}) = 54

What is the answer to 2^x+2^{-x}=5 ?
The answer to 2^x+2^{-x}=5 is x=(ln((5+sqrt(21))/2))/(ln(2)),x=(ln((5-sqrt(21))/2))/(ln(2))
Solve for x: 2^x+2^{-x}=5
The solution is x=(ln((5+sqrt(21))/2))/(ln(2)),x=(ln((5-sqrt(21))/2))/(ln(2))