What is the answer to g(e^n-1)=ln(e^n-1-1) ?

The answer to g(e^n-1)=ln(e^n-1-1) is n=ln(-(\W(-ge^g)-2g)/g)

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g(e^n-1)=ln(e^n-1-1)

g (e^{n} - 1) = ln (e^{n} - 1 - 1)

n = ln (- \frac{W ( - g e ^{g} ) - 2 g}{g})

Solution steps

g (e^{n} - 1) = ln (e^{n} - 1 - 1)

Rewrite the equation with

e^{n} = u

g (u - 1) = ln (u - 1 - 1)

Solve

g (u - 1) = ln (u - 1 - 1) : u = - \frac{W ( - g e ^{g} ) - 2 g}{g}

u = - \frac{W ( - g e ^{g} ) - 2 g}{g}

Substitute back

u = e^{n},

solve for

n

Solve

e^{n} = - \frac{W ( - g e ^{g} ) - 2 g}{g} : n = ln (- \frac{W ( - g e ^{g} ) - 2 g}{g})

n = ln (- \frac{W ( - g e ^{g} ) - 2 g}{g})

2^{x} = e^{5}

5^{5 x + 1} = 12 5^{x}

3^{x + 2} = 1

3^{x + 2} = 6

3 2^{x - 1} = 64

What is the answer to g(e^n-1)=ln(e^n-1-1) ?
The answer to g(e^n-1)=ln(e^n-1-1) is n=ln(-(\W(-ge^g)-2g)/g)