Solve #(logx^2)^2 - logx^3 - 10 = 0# for #x#?

Question

All bases are equal (2).

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Answer 1 · 2017-09-22T11:09:21

#x=4 or approx 0.420448#

#(logx^2)^2 - logx^3-10 = 0#

Given #log = log_2#

#(2logx)^2-3logx-10=0#

#4*(logx)^2-3logx-10=0#

Let #lambda = logx -> x=2^lambda#

#4lambda^2-3lambda-10 =0#

#(4lambda+5)(lambda-2)=0#

#lambda = -5/4 or +2#

#lambda = 2 -> x=2^2 =4#

#lambda = -5/4 -> x= 2^(-5/4) = 1/root4 32 approx 0.420448#

Hence, #x=4 or approx 0.420448#