Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$${solve}\:\:\frac{\mathrm{1}}{\:\sqrt{{x}}}={t}\Rightarrow\frac{\mathrm{1}}{{x}}={t}^{\mathrm{2}} \Rightarrow{x}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$$\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }×\frac{\mathrm{1}}{{t}}} +\mathrm{3}^{\mathrm{1}} ×\mathrm{3}^{{t}} =\mathrm{12} \\ $$$$\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }} +\mathrm{3}^{\mathrm{1}} ×\mathrm{3}^{{t}} =\mathrm{12} \\ $$$$\mathrm{3}^{{t}} \left(\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}} +\mathrm{3}\right)=\left(\mathrm{3}×\mathrm{4}\right) \\ $$$$\mathrm{3}^{{t}} =\mathrm{3}^{\mathrm{1}} \Rightarrow{t}=\mathrm{1} \\ $$$$\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}} =\mathrm{1}\Rightarrow\mathrm{3}^{\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}} =\mathrm{3}^{\mathrm{0}} \\ $$$$\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}=\mathrm{0}\Rightarrow\mathrm{1}−{t}^{\mathrm{4}} =\mathrm{0}\Rightarrow{t}^{\mathrm{4}} =\mathrm{1}\Rightarrow{t}=\mathrm{1} \\ $$$${how}\:\:\:{x}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\Rightarrow{x}=\mathrm{1} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com