What is the answer to 2^x-2^{-x}=20 ?

The answer to 2^x-2^{-x}=20 is x=(ln(10+sqrt(101)))/(ln(2))

Solve for x: 2^x-2^{-x}=20

The solution is x=(ln(10+sqrt(101)))/(ln(2))

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2^{x} - 2^{- x} = 20

x = \frac{ln ( 10 + 101 )}{ln ( 2 )}

Solution steps

2^{x} - 2^{- x} = 20

Apply exponent rules

2^{x} - (2^{x})^{- 1} = 20

Rewrite the equation with

2^{x} = u

u - (u)^{- 1} = 20

Solve

u - u^{- 1} = 20 : u = 10 + 101, u = 10 - 101

u = 10 + 101, u = 10 - 101

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 10 + 101 : x = \frac{ln ( 10 + 101 )}{ln ( 2 )}

Solve

2^{x} = 10 - 101 :

No Solution for

x \in R

x = \frac{ln ( 10 + 101 )}{ln ( 2 )}

3^{2 x + 1} = 5

x, 2^{x + y} = 8

97 \cdot 2^{2 x} = 69

3^{9 x} = 3^{7 x + 8}

2000 = 4000 e^{- 0.1 x}

What is the answer to 2^x-2^{-x}=20 ?
The answer to 2^x-2^{-x}=20 is x=(ln(10+sqrt(101)))/(ln(2))
Solve for x: 2^x-2^{-x}=20
The solution is x=(ln(10+sqrt(101)))/(ln(2))