Solve ${({log x}^{2})}^{2} - {log x}^{3} - 10 = 0$ $(logx^2)^2 - logx^3 - 10 = 0$ for $x$ $x$ ?

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All bases are equal (2).

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Answer 1 · 2017-09-22T11:09:21

$x = 4 or \approx 0.420448$ $x=4 or approx 0.420448$

${({log x}^{2})}^{2} - {log x}^{3} - 10 = 0$ $(logx^2)^2 - logx^3-10 = 0$

Given $log = {log}_{2}$ $log = log_2$

${(2 log x)}^{2} - 3 log x - 10 = 0$ $(2logx)^2-3logx-10=0$

$4 \cdot {(log x)}^{2} - 3 log x - 10 = 0$ $4*(logx)^2-3logx-10=0$

Let $λ = log x \to x = 2^{λ}$ $lambda = logx -> x=2^lambda$

$4 λ^{2} - 3 λ - 10 = 0$ $4lambda^2-3lambda-10 =0$

$(4 λ + 5) (λ - 2) = 0$ $(4lambda+5)(lambda-2)=0$

$λ = - \frac{5}{4} or + 2$ $lambda = -5/4 or +2$

$λ = 2 \to x = 2^{2} = 4$ $lambda = 2 -> x=2^2 =4$

$λ = - \frac{5}{4} \to x = 2^{- \frac{5}{4}} = \frac{1}{\sqrt[4]{32}} \approx 0.420448$ $lambda = -5/4 -> x= 2^(-5/4) = 1/root4 32 approx 0.420448$

Hence, $x = 4 or \approx 0.420448$ $x=4 or approx 0.420448$