What is the answer to 3^x-3^{-x}+2=0 ?

The answer to 3^x-3^{-x}+2=0 is x=(ln(-1+sqrt(2)))/(ln(3))

Solve for x: 3^x-3^{-x}+2=0

The solution is x=(ln(-1+sqrt(2)))/(ln(3))

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

x = \frac{ln ( - 1 + 2 )}{ln ( 3 )}

Decimal Notation

x = - 0.80226 \dots

Solution steps

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

Apply exponent rules

3^{x} - (3^{x})^{- 1} + 2 = 0

Rewrite the equation with

3^{x} = u

u - (u)^{- 1} + 2 = 0

Solve

u - u^{- 1} + 2 = 0 : u = - 1 + 2, u = - 1 - 2

u = - 1 + 2, u = - 1 - 2

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = - 1 + 2 : x = \frac{ln ( - 1 + 2 )}{ln ( 3 )}

Solve

3^{x} = - 1 - 2 :

No Solution for

x \in R

x = \frac{ln ( - 1 + 2 )}{ln ( 3 )}

10 x - 20.89 \cdot e^{\frac{- x}{( 2478.82 )}} = 0

4^{x} = 5^{x + 1}

3^{x^{2} - 1} = 134

x^{x} = 100

y, e^{3 x} + x^{2} \cdot y + e^{y} = 0

What is the answer to 3^x-3^{-x}+2=0 ?
The answer to 3^x-3^{-x}+2=0 is x=(ln(-1+sqrt(2)))/(ln(3))
Solve for x: 3^x-3^{-x}+2=0
The solution is x=(ln(-1+sqrt(2)))/(ln(3))