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)])}#