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Solve-x-R-x2-1-x-1-x-2-x-4-Source-L-Panaitopol-

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Question Number 173485 by mnjuly1970 last updated on 12/Jul/22
Solve (x ∈ R ) x2^( (1/x)) +(1/x) 2^( x) = 4 −Source: L.Panaitopol
$$ \\ $$$$\:\:\:\:\:{Solve}\:\:\:\:\:\:\:\left({x}\:\in\:\mathbb{R}\:\right) \\ $$$$\:\:\:\:\:\:{x}\mathrm{2}^{\:\frac{\mathrm{1}}{{x}}} \:+\frac{\mathrm{1}}{{x}}\:\mathrm{2}^{\:{x}} =\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:−{Source}:\:{L}.{Panaitopol} \\ $$
Answered by aleks041103 last updated on 12/Jul/22
obv x>0 for x2^(1/x) +2^x /x = 4 f(x)=x 2^(1/x) ⇒g(x)=f(x)+f(1/x) ⇒g(1/x)=f(1/x)+f(x) ⇒g(x)=g(1/x)=4 ⇒therefore it is enough to search for x>1 and for ∀x_i >1, s.t. g(x_i )=4 we′ll have x_i ^� =(1/x_i ), s.t. g(x_i ^� )=4 too. 2^x =e^(ln(2)x) =Σ_(i=0) ^∞ (((ln(2)x)^i )/(i!)) ⇒f(x)=Σ_(i=0) ^∞ (((ln(2))^i )/(i!)) (1/x^(i−1) ) ⇒f(1/x)=Σ_(i=0) ^∞ (((ln(2))^i )/(i!))x^(i−1) ⇒g(x)=Σ_(i=0) ^∞ (((ln(2))^i )/(i!))(x^(i−1) +(1/x^(i−1) )) g(x)=Σ_(i=0) ^∞ (((ln(2))^i )/(i!))s_(i−1) (x) s_i (x)=x^i +x^(−i) s_i ′(x)=ix^(i−1) −ix^(−i−1) =(i/x)(x^i −x^(−i) ) for x>1: i>0:x^i >1⇒x^(−i) <1⇒x^i −x^(−i) >0 ⇒s_(i>0) ′(x>1)>0 i<0:x^(i<1) ⇒x^(−i) >1⇒x^i −x^(−i) <0 ⇒i(x^i −x^(−i) )>0 ⇒s_(i<0) ′(x>1)>0 i=0: s_0 ′=0 ⇒g′(x>1)=Σ_(i=0,i≠1) ^∞ (((ln(2))^i )/(i!))s_(i−1) ′(x>1) since s_(i−1) ′(x>1)>0⇒g′(x>1)>0 ⇒g(x) is strictly increasing for x>1 ⇒for g(x)=4 exists at most 1 soln. x≥1 And g(1)=4 ⇒The only solution to the problem is x=1
$${obv}\:{x}>\mathrm{0}\:{for}\:{x}\mathrm{2}^{\mathrm{1}/{x}} +\mathrm{2}^{{x}} /{x}\:=\:\mathrm{4} \\ $$$${f}\left({x}\right)={x}\:\mathrm{2}^{\mathrm{1}/{x}} \\ $$$$\Rightarrow{g}\left({x}\right)={f}\left({x}\right)+{f}\left(\mathrm{1}/{x}\right) \\ $$$$\Rightarrow{g}\left(\mathrm{1}/{x}\right)={f}\left(\mathrm{1}/{x}\right)+{f}\left({x}\right) \\ $$$$\Rightarrow{g}\left({x}\right)={g}\left(\mathrm{1}/{x}\right)=\mathrm{4} \\ $$$$\Rightarrow{therefore}\:{it}\:{is}\:{enough}\:{to}\:{search} \\ $$$${for}\:{x}>\mathrm{1}\:{and}\:{for}\:\forall{x}_{{i}} >\mathrm{1},\:{s}.{t}.\:{g}\left({x}_{{i}} \right)=\mathrm{4}\:{we}'{ll} \\ $$$${have}\:\bar {{x}}_{{i}} =\frac{\mathrm{1}}{{x}_{{i}} },\:{s}.{t}.\:{g}\left(\bar {{x}}_{{i}} \right)=\mathrm{4}\:{too}. \\ $$$$\mathrm{2}^{{x}} ={e}^{{ln}\left(\mathrm{2}\right){x}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({ln}\left(\mathrm{2}\right){x}\right)^{{i}} }{{i}!} \\ $$$$\Rightarrow{f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({ln}\left(\mathrm{2}\right)\right)^{{i}} }{{i}!}\:\frac{\mathrm{1}}{{x}^{{i}−\mathrm{1}} } \\ $$$$\Rightarrow{f}\left(\mathrm{1}/{x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({ln}\left(\mathrm{2}\right)\right)^{{i}} }{{i}!}{x}^{{i}−\mathrm{1}} \\ $$$$\Rightarrow{g}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({ln}\left(\mathrm{2}\right)\right)^{{i}} }{{i}!}\left({x}^{{i}−\mathrm{1}} +\frac{\mathrm{1}}{{x}^{{i}−\mathrm{1}} }\right) \\ $$$${g}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({ln}\left(\mathrm{2}\right)\right)^{{i}} }{{i}!}{s}_{{i}−\mathrm{1}} \left({x}\right) \\ $$$${s}_{{i}} \left({x}\right)={x}^{{i}} +{x}^{−{i}} \\ $$$${s}_{{i}} '\left({x}\right)={ix}^{{i}−\mathrm{1}} −{ix}^{−{i}−\mathrm{1}} =\frac{{i}}{{x}}\left({x}^{{i}} −{x}^{−{i}} \right) \\ $$$${for}\:{x}>\mathrm{1}: \\ $$$${i}>\mathrm{0}:{x}^{{i}} >\mathrm{1}\Rightarrow{x}^{−{i}} <\mathrm{1}\Rightarrow{x}^{{i}} −{x}^{−{i}} >\mathrm{0} \\ $$$$\Rightarrow{s}_{{i}>\mathrm{0}} '\left({x}>\mathrm{1}\right)>\mathrm{0} \\ $$$${i}<\mathrm{0}:{x}^{{i}<\mathrm{1}} \Rightarrow{x}^{−{i}} >\mathrm{1}\Rightarrow{x}^{{i}} −{x}^{−{i}} <\mathrm{0} \\ $$$$\Rightarrow{i}\left({x}^{{i}} −{x}^{−{i}} \right)>\mathrm{0} \\ $$$$\Rightarrow{s}_{{i}<\mathrm{0}} '\left({x}>\mathrm{1}\right)>\mathrm{0} \\ $$$${i}=\mathrm{0}: \\ $$$${s}_{\mathrm{0}} '=\mathrm{0} \\ $$$$\Rightarrow{g}'\left({x}>\mathrm{1}\right)=\underset{{i}=\mathrm{0},{i}\neq\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({ln}\left(\mathrm{2}\right)\right)^{{i}} }{{i}!}{s}_{{i}−\mathrm{1}} '\left({x}>\mathrm{1}\right) \\ $$$${since}\:{s}_{{i}−\mathrm{1}} '\left({x}>\mathrm{1}\right)>\mathrm{0}\Rightarrow{g}'\left({x}>\mathrm{1}\right)>\mathrm{0} \\ $$$$\Rightarrow{g}\left({x}\right)\:{is}\:{strictly}\:{increasing}\:{for}\:{x}>\mathrm{1} \\ $$$$\Rightarrow{for}\:{g}\left({x}\right)=\mathrm{4}\:{exists}\:{at}\:{most}\:\mathrm{1}\:{soln}.\:{x}\geqslant\mathrm{1} \\ $$$${And}\:{g}\left(\mathrm{1}\right)=\mathrm{4} \\ $$$$\Rightarrow{The}\:{only}\:{solution}\:{to}\:{the}\:{problem}\:{is} \\ $$$${x}=\mathrm{1} \\ $$
Commented by Tawa11 last updated on 13/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by dragan91 last updated on 12/Jul/22
2^(log_2 x+(1/x)) +2^(log_2 (1/x)+x) =4 2^(log_2 x+(1/x)) +2^(−log_2 x+x) =4 2^(log_2 x+(1/x)) +2^(−log_2 x+x ) ≥^(AM−GM) 2(√2^(x+(1/x)) ) 2^(x+(1/x)) ≤2^2 ∣log_2 x+(1/x)≤2 since x+(1/x)≥^(AM−GM) 2⇒equation has solution only for x=1
$$\mathrm{2}^{\mathrm{log}_{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}} +\mathrm{2}^{\mathrm{log}_{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{x}}+\mathrm{x}} =\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{log}_{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}} +\mathrm{2}^{−\mathrm{log}_{\mathrm{2}} \mathrm{x}+\mathrm{x}} =\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{log}_{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}} +\mathrm{2}^{−\mathrm{log}_{\mathrm{2}} \mathrm{x}+\mathrm{x}\:\:} \overset{\mathrm{AM}−\mathrm{GM}} {\geqslant}\:\:\mathrm{2}\sqrt{\mathrm{2}^{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}} } \\ $$$$\mathrm{2}^{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}} \leqslant\mathrm{2}^{\mathrm{2}} \mid\mathrm{log}_{\mathrm{2}} \\ $$$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\leqslant\mathrm{2} \\ $$$$\mathrm{since}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\overset{\mathrm{AM}−\mathrm{GM}} {\geqslant}\mathrm{2}\Rightarrow\mathrm{equation}\:\mathrm{has}\:\mathrm{solution}\: \\ $$$$\mathrm{only}\:\mathrm{for}\:\mathrm{x}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 13/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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