What is the answer to solvefor x,y^-=ln(e^{2x}-1) ?

The answer to solvefor x,y^-=ln(e^{2x}-1) is x=(ln(e^{y^-}+1))/2

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solvefor x,y^-=ln(e^{2x}-1)

solve for

x, y^{-} = ln (e^{2 x} - 1)

x = \frac{ln ( e ^{y^{-}} + 1 )}{2}

Solution steps

y^{-} = ln (e^{2 x} - 1)

Rewrite the equation with

e^{2 x} = u

y^{-} = ln (u - 1)

Solve

y^{-} = ln (u - 1) : u = e^{y^{-}} + 1

u = e^{y^{-}} + 1

Substitute back

u = e^{2 x},

solve for

x

Solve

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

x = \frac{ln ( e ^{y^{-}} + 1 )}{2}

3^{x^{2} - x - 12} = 7^{x^{2} - x - 12}

e^{6 x} - e^{3 x} - 2 = 0

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

9^{8 x - 4} = \frac{1}{( 81 ) ^{2 x - 5}}

\frac{1}{4 ^{x - 2}} = 64

What is the answer to solvefor x,y^-=ln(e^{2x}-1) ?
The answer to solvefor x,y^-=ln(e^{2x}-1) is x=(ln(e^{y^-}+1))/2