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

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

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

solve for

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

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

Solution steps

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

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Substitute back

u = e^{x},

solve for

x

Solve

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

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

e^{l n (x)} = 0

2^{x + 1} - 2^{x - 3} + 2^{x} = 23

4^{x} + 2^{x + 3} = 48

x, e^{x} + 5 e^{- x} = 6

n, 2^{n} = 1000000

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