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Question Number 101243 by Dwaipayan Shikari last updated on 01/Jul/20

Find the solution     xa^(1/x) +(1/x)a^x =2a   a∈{−1,0,1}   and also find when a  is not given

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{solution}\: \\ $$$$\:\:\mathrm{xa}^{\frac{\mathrm{1}}{\mathrm{x}}} +\frac{\mathrm{1}}{\mathrm{x}}\mathrm{a}^{\mathrm{x}} =\mathrm{2a}\:\:\:\mathrm{a}\in\left\{−\mathrm{1},\mathrm{0},\mathrm{1}\right\}\:\:\:{and}\:{also}\:{find}\:{when}\:{a}\:\:{is}\:{not}\:{given} \\ $$

Commented by mr W last updated on 01/Jul/20

for a>0:  x=1 is the only solution.

$${for}\:{a}>\mathrm{0}: \\ $$$${x}=\mathrm{1}\:{is}\:{the}\:{only}\:{solution}. \\ $$

Commented by mr W last updated on 01/Jul/20

a=−1:  x(−1)^(1/x) +(((−1)^x )/x)=−2  ⇒x=1    a=0:  ⇒x∈R−{0}    a=1:  x+(1/x)=2  ⇒x=1

$${a}=−\mathrm{1}: \\ $$$${x}\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{{x}}} +\frac{\left(−\mathrm{1}\right)^{{x}} }{{x}}=−\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$$ \\ $$$${a}=\mathrm{0}: \\ $$$$\Rightarrow{x}\in{R}−\left\{\mathrm{0}\right\} \\ $$$$ \\ $$$${a}=\mathrm{1}: \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$

Commented by Dwaipayan Shikari last updated on 01/Jul/20

You are right sir. Can you find when a is not given

$${You}\:{are}\:{right}\:{sir}.\:{Can}\:{you}\:{find}\:{when}\:{a}\:{is}\:{not}\:{given} \\ $$

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