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, $x=4 or approx 0.420448$