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Question Number 74244 by Maclaurin Stickker last updated on 20/Nov/19
Find x x^(log_(4 ) (√x)) =x^(log_4 x) −2
$${Find}\:{x} \\ $$$${x}^{{log}_{\mathrm{4}\:} \sqrt{{x}}} ={x}^{{log}_{\mathrm{4}} {x}} −\mathrm{2} \\ $$
Answered by MJS last updated on 20/Nov/19
t=x^(log_4 (√x)) =x^((ln (√x))/(ln 4)) =x^((ln x)/(4ln 2)) ln t =((ln x)/(4ln 2))ln x =(((ln x)^2 )/(4ln 2)) ⇒ x=e^(2(√(ln 2 ln t))) x^(log_4 (√x)) =t x^(log_4 x) =t^2 t=t^2 −2 t^2 −t−2=0 ⇒ t=−1∨t=2 but for x∈R we need t>1 ⇒ t=2 ⇒ x=4
$${t}={x}^{\mathrm{log}_{\mathrm{4}} \:\sqrt{{x}}} ={x}^{\frac{\mathrm{ln}\:\sqrt{{x}}}{\mathrm{ln}\:\mathrm{4}}} ={x}^{\frac{\mathrm{ln}\:{x}}{\mathrm{4ln}\:\mathrm{2}}} \\ $$$$\mathrm{ln}\:{t}\:=\frac{\mathrm{ln}\:{x}}{\mathrm{4ln}\:\mathrm{2}}\mathrm{ln}\:{x}\:=\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{4ln}\:\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\mathrm{e}^{\mathrm{2}\sqrt{\mathrm{ln}\:\mathrm{2}\:\mathrm{ln}\:{t}}} \\ $$$${x}^{\mathrm{log}_{\mathrm{4}} \:\sqrt{{x}}} ={t} \\ $$$${x}^{\mathrm{log}_{\mathrm{4}} \:{x}} ={t}^{\mathrm{2}} \\ $$$${t}={t}^{\mathrm{2}} −\mathrm{2} \\ $$$${t}^{\mathrm{2}} −{t}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:{t}=−\mathrm{1}\vee{t}=\mathrm{2}\:\mathrm{but}\:\mathrm{for}\:{x}\in\mathbb{R}\:\mathrm{we}\:\mathrm{need}\:{t}>\mathrm{1} \\ $$$$\Rightarrow\:{t}=\mathrm{2}\:\Rightarrow\:{x}=\mathrm{4} \\ $$
Answered by mr W last updated on 20/Nov/19
x>0 let u=x^(log_4 (√x)) =x^((1/2)log_4 x) =(√x^(log_4 x) )>0 ⇒u=u^2 −2 ⇒u^2 −u−2=0 ⇒(u−2)(u+1)=0 ⇒u=2, −1 with u=2: (√x^(log_4 x) )=2 x^(log_4 x) =4 x^((ln x)/(ln 4)) =4 ((ln x)/(ln 4))=((ln 4)/(ln x)) (ln x)^2 =(ln 4)^2 ln x=±ln 4=ln 4^(±1) ⇒x=4^(±1) =4 or (1/4)
$${x}>\mathrm{0} \\ $$$${let}\:{u}={x}^{\mathrm{log}_{\mathrm{4}} \:\sqrt{{x}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mathrm{4}} \:{x}} =\sqrt{{x}^{\mathrm{log}_{\mathrm{4}} \:{x}} }>\mathrm{0} \\ $$$$\Rightarrow{u}={u}^{\mathrm{2}} −\mathrm{2} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −{u}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({u}−\mathrm{2}\right)\left({u}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{2},\:−\mathrm{1} \\ $$$${with}\:{u}=\mathrm{2}: \\ $$$$\sqrt{{x}^{\mathrm{log}_{\mathrm{4}} \:{x}} }=\mathrm{2} \\ $$$${x}^{\mathrm{log}_{\mathrm{4}} \:{x}} =\mathrm{4} \\ $$$${x}^{\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{4}}} =\mathrm{4} \\ $$$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{4}}=\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{ln}\:{x}} \\ $$$$\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} =\left(\mathrm{ln}\:\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\mathrm{ln}\:{x}=\pm\mathrm{ln}\:\mathrm{4}=\mathrm{ln}\:\mathrm{4}^{\pm\mathrm{1}} \\ $$$$\Rightarrow{x}=\mathrm{4}^{\pm\mathrm{1}} =\mathrm{4}\:{or}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by MJS last updated on 20/Nov/19
same path we share... but the given equation also has the solution x=(1/4) how to get it?
$$\mathrm{same}\:\mathrm{path}\:\mathrm{we}\:\mathrm{share}… \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{also}\:\mathrm{has}\:\mathrm{the}\:\mathrm{solution} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{it}? \\ $$
Commented by mr W last updated on 20/Nov/19
two solutions indeed. see above.
$${two}\:{solutions}\:{indeed}.\:{see}\:{above}. \\ $$
Commented by Maclaurin Stickker last updated on 21/Nov/19
Sir, I added 2 by the sides and applied log_2 in both sides. Is this correct? I got the same results
$${Sir},\:{I}\:{added}\:\mathrm{2}\:{by}\:{the}\:{sides}\:{and}\:{applied} \\ $$$${log}_{\mathrm{2}} \:{in}\:{both}\:{sides}.\:{Is}\:{this}\:{correct}? \\ $$$${I}\:{got}\:{the}\:{same}\:{results} \\ $$
Commented by MJS last updated on 21/Nov/19
doesn′t sound correct to me, please post it
$$\mathrm{doesn}'\mathrm{t}\:\mathrm{sound}\:\mathrm{correct}\:\mathrm{to}\:\mathrm{me},\:\mathrm{please}\:\mathrm{post}\:\mathrm{it} \\ $$
Commented by Maclaurin Stickker last updated on 21/Nov/19
x^(log_4 (√x)) +2=x^(log_4 x) log_2 (x^(log_4 (√x)) )+log_2 2=log_2 (x^(log_4 x) ) .... (1/4)log_2 x^2 =(1/2)log_2 x^2 −1 .... log_2 x^2 =4 log_2 x=2 and log_2 x=−2 x=4 or x=(1/4)
$${x}^{{log}_{\mathrm{4}} \sqrt{{x}}} +\mathrm{2}={x}^{{log}_{\mathrm{4}} {x}} \\ $$$${log}_{\mathrm{2}} \left({x}^{{log}_{\mathrm{4}} \sqrt{{x}}} \right)+{log}_{\mathrm{2}} \mathrm{2}={log}_{\mathrm{2}} \left({x}^{{log}_{\mathrm{4}} {x}} \right) \\ $$$$…. \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{log}_{\mathrm{2}} {x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{log}_{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{1} \\ $$$$…. \\ $$$${log}_{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{4} \\ $$$${log}_{\mathrm{2}} {x}=\mathrm{2}\:\:{and}\:\:{log}_{\mathrm{2}} {x}=−\mathrm{2} \\ $$$${x}=\mathrm{4}\:\:{or}\:{x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by MJS last updated on 21/Nov/19
funny you get the solution... but log_b (p+q) ≠log_b p +log_b q b^(log_b (p+q)) =p+a but b^(log_b p +log_b q) =pq so your 2^(nd) line is not equivalent to the first line
$$\mathrm{funny}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{solution}… \\ $$$$\mathrm{but} \\ $$$$\mathrm{log}_{{b}} \:\left({p}+{q}\right)\:\neq\mathrm{log}_{{b}} \:{p}\:+\mathrm{log}_{{b}} \:{q} \\ $$$${b}^{\mathrm{log}_{{b}} \:\left({p}+{q}\right)} ={p}+{a}\:\mathrm{but}\:{b}^{\mathrm{log}_{{b}} \:{p}\:+\mathrm{log}_{{b}} \:{q}} ={pq} \\ $$$$\mathrm{so}\:\mathrm{your}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{line}\:\mathrm{is}\:\mathrm{not}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{line} \\ $$

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