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

The answer to solvefor r,y=r(x-1)^r+1 is r=(\W(yln(x-1)-ln(x-1)))/(ln(x-1))

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solvefor r,y=r(x-1)^r+1

solve for

r, y = r (x - 1)^{r} + 1

r = \frac{W ( y ln ( x - 1 ) - ln ( x - 1 ) )}{ln ( x - 1 )}

Solution steps

y = r (x - 1)^{r} + 1

Prepare

y = r (x - 1)^{r} + 1

for Lambert form:

r e^{l n (x - 1) r} = y - 1

Rewrite the equation with

r ln (x - 1) = u

and

r = \frac{u}{ln ( x - 1 )}

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

Rewrite

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

in Lambert form:

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

Solve

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

u = W (y ln (x - 1) - ln (x - 1))

Substitute back

u = r ln (x - 1),

solve for

r

Solve

r ln (x - 1) = W (y ln (x - 1) - ln (x - 1)) : r = \frac{W ( y ln ( x - 1 ) - ln ( x - 1 ) )}{ln ( x - 1 )}

r = \frac{W ( y ln ( x - 1 ) - ln ( x - 1 ) )}{ln ( x - 1 )}

2^{x^{2}} - 4^{3 x - 4} = 0

(x) = 3^{- x}

x, x + y + e^{x y} = 0

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

What is the answer to solvefor r,y=r(x-1)^r+1 ?
The answer to solvefor r,y=r(x-1)^r+1 is r=(\W(yln(x-1)-ln(x-1)))/(ln(x-1))