What is the answer to solvefor x,e^y=ln(1+e^x) ?

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

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solvefor x,e^y=ln(1+e^x)

solve for

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

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

Solution steps

e^{y} = ln (1 + e^{x})

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Substitute back

u = e^{x},

solve for

x

Solve

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

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

5^{x} \cdot 2 5^{2 x - 1} = 1

1250 = 10 (5)^{x}

9^{2 x + 5} = 1 7^{x + 8}

\frac{( e ^{x + 5} )}{e ^{5}} = 3

e^{9 - 4 x} = 7

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