How do you solve $(x-1)^[log(x-1)]=100(x-1)$ ?

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Answer 1 · 2016-06-18T23:52:56

$x = 1 + e^{1/2 (1 + sqrt[1 + 4 Log_e(100)])}$

Making $y = x-1$ we have

$y^{log_e y} = 100 y->y^{log_e y-1} = 100$

Applying $log$ to both sides

$(log_e y - 1)log_e y = log_e100$

Solving for $log_e y$ we have

$log_e y = 1/2 (1 pm sqrt[1 + 4 Log_e(100)])$

so

$Y = x-1=e^{1/2 (1 pm sqrt[1 + 4 Log_e(100)])}$

Finally considering that $x -1> 0$

$x = 1 + e^{1/2 (1 + sqrt[1 + 4 Log_e(100)])}$

How do you solve (x-1)^[log(x-1)]=100(x-1)(x-1)^[log(x-1)]=100(x-1)?