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Question Number 163452 by HongKing last updated on 07/Jan/22
Solve for real numbers: 3^(x (√x)) + 3^(1 + (1/( (√x)))) = 12
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{3}^{\boldsymbol{\mathrm{x}}\:\sqrt{\boldsymbol{\mathrm{x}}}} \:\:+\:\:\mathrm{3}^{\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}}}} \:\:=\:\mathrm{12} \\ $$
Commented by mr W last updated on 07/Jan/22
i don′t think such equations have much technique values. basically we must “see” or guess the solutions. here for example: we guess 3^2 +3=12 then see x=1 is ok. i say this has no technique value, but rather stupid, because we are dead if the equation is 3^(x (√x)) + 3^(1 + (1/( (√x)))) = 11. i have never understood what the creators (i don′t mean Hongking sir) of such questions want to test from the students.
$${i}\:{don}'{t}\:{think}\:{such}\:{equations}\:{have} \\ $$$${much}\:{technique}\:{values}.\:{basically} \\ $$$${we}\:{must}\:“{see}''\:{or}\:{guess}\:{the}\:{solutions}. \\ $$$${here}\:{for}\:{example}:\:{we}\:{guess}\:\mathrm{3}^{\mathrm{2}} +\mathrm{3}=\mathrm{12} \\ $$$${then}\:{see}\:{x}=\mathrm{1}\:{is}\:{ok}. \\ $$$${i}\:{say}\:{this}\:{has}\:{no}\:{technique}\:{value}, \\ $$$${but}\:{rather}\:{stupid},\:{because}\:{we}\:{are}\:{dead} \\ $$$${if}\:{the}\:{equation}\:{is} \\ $$$$\mathrm{3}^{\boldsymbol{\mathrm{x}}\:\sqrt{\boldsymbol{\mathrm{x}}}} \:\:+\:\:\mathrm{3}^{\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}}}} \:\:=\:\mathrm{11}. \\ $$$${i}\:{have}\:{never}\:{understood}\:{what}\:{the} \\ $$$${creators}\:\left({i}\:{don}'{t}\:{mean}\:{Hongking}\:{sir}\right) \\ $$$${of}\:{such}\:{questions}\:{want}\:{to}\:{test}\:{from} \\ $$$${the}\:{students}. \\ $$
Answered by mathlove last updated on 08/Jan/22
solve (1/( (√x)))=t⇒(1/x)=t^2 ⇒x=(1/t^2 ) 3^((1/t^2 )×(1/t)) +3^1 ×3^t =12 3^(1/t^3 ) +3^1 ×3^t =12 3^t (3^((1/t^3 )−t) +3)=(3×4) 3^t =3^1 ⇒t=1 3^((1/t^3 )−t) =1⇒3^((1/t^3 )−t) =3^0 (1/t^3 )−t=0⇒1−t^4 =0⇒t^4 =1⇒t=1 how x=(1/t^2 )⇒x=1
$${solve}\:\:\frac{\mathrm{1}}{\:\sqrt{{x}}}={t}\Rightarrow\frac{\mathrm{1}}{{x}}={t}^{\mathrm{2}} \Rightarrow{x}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$$\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }×\frac{\mathrm{1}}{{t}}} +\mathrm{3}^{\mathrm{1}} ×\mathrm{3}^{{t}} =\mathrm{12} \\ $$$$\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }} +\mathrm{3}^{\mathrm{1}} ×\mathrm{3}^{{t}} =\mathrm{12} \\ $$$$\mathrm{3}^{{t}} \left(\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}} +\mathrm{3}\right)=\left(\mathrm{3}×\mathrm{4}\right) \\ $$$$\mathrm{3}^{{t}} =\mathrm{3}^{\mathrm{1}} \Rightarrow{t}=\mathrm{1} \\ $$$$\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}} =\mathrm{1}\Rightarrow\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}} =\mathrm{3}^{\mathrm{0}} \\ $$$$\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}=\mathrm{0}\Rightarrow\mathrm{1}−{t}^{\mathrm{4}} =\mathrm{0}\Rightarrow{t}^{\mathrm{4}} =\mathrm{1}\Rightarrow{t}=\mathrm{1} \\ $$$${how}\:\:\:{x}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\Rightarrow{x}=\mathrm{1} \\ $$$$ \\ $$

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