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

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Answer 1 · 2016-06-18T23:52:56

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

Making $y = x - 1$ $y = x-1$ we have

$y^{{log}_{e} y} = 100 y \to y^{{log}_{e} y - 1} = 100$ $y^{log_e y} = 100 y->y^{log_e y-1} = 100$

Applying $log$ $log$ to both sides

$({log}_{e} y - 1) {log}_{e} y = {log}_{e} 100$ $(log_e y - 1)log_e y = log_e100$

Solving for ${log}_{e} y$ $log_e y$ we have

${log}_{e} y = \frac{1}{2} (1 \pm \sqrt{1 + 4 {log}_{e} (100)})$ $log_e y = 1/2 (1 pm sqrt[1 + 4 Log_e(100)])$

so

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

Finally considering that $x - 1 > 0$ $x -1> 0$

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

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