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Question Number 91177 by jagoll last updated on 28/Apr/20

6^((log_2  x)^2 )  + x^((log_2  x))  = 12

$$\mathrm{6}^{\left(\mathrm{log}_{\mathrm{2}} \:{x}\right)^{\mathrm{2}} } \:+\:{x}^{\left(\mathrm{log}_{\mathrm{2}} \:{x}\right)} \:=\:\mathrm{12}\: \\ $$

Commented by jagoll last updated on 28/Apr/20

i have no idea to solve this

$${i}\:{have}\:{no}\:{idea}\:{to}\:{solve}\:{this} \\ $$

Answered by MJS last updated on 28/Apr/20

let t=e^(((ln x)^2 )/(ln 2))  ⇔ x=e^(±(√(ln 2 ln t)))   t^(1+((ln 3)/(ln 2))) +t=12  I don′t think we can solve exactly  t≈2.39890750  ⇒  x≈.458961247∨x≈2.17883320

$$\mathrm{let}\:{t}=\mathrm{e}^{\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{ln}\:\mathrm{2}}} \:\Leftrightarrow\:{x}=\mathrm{e}^{\pm\sqrt{\mathrm{ln}\:\mathrm{2}\:\mathrm{ln}\:{t}}} \\ $$$${t}^{\mathrm{1}+\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}} +{t}=\mathrm{12} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{exactly} \\ $$$${t}\approx\mathrm{2}.\mathrm{39890750} \\ $$$$\Rightarrow \\ $$$${x}\approx.\mathrm{458961247}\vee{x}\approx\mathrm{2}.\mathrm{17883320} \\ $$

Commented by jagoll last updated on 28/Apr/20

wow..an uncomfortable question

$${wow}..{an}\:{uncomfortable}\:{question} \\ $$

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