Solve-for-x-7x-1-3-x-x-0-

Question Number 221620 by fantastic last updated on 08/Jun/25

$${Solve}\:{for}\:{x} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{7}{x}}=\sqrt{{x}}\left[{x}\neq\mathrm{0}\right] \\ $$

Answered by fantastic last updated on 08/Jun/25

or (7x)^(1/3) =x^(1/2) or (7)^(1/3) .x^(1/3) =x^(1/2) or (7)^(1/3) =(x^(1/2) /x^(1/3) ) or (7)^(1/3) =x^(((1/2)−(1/3))) or (7)^(1/3) =x^((((3−2)/6))) or (7)^(1/3) =x^(1/6) or 7^(1/3) =x^(1/6) or 7^(6/3) =x^(6/6) or 7^2 =x^1 So x=49

$${or}\:\left(\mathrm{7}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${or}\:\sqrt[{\mathrm{3}}]{\mathrm{7}}.{x}^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${or}\:\sqrt[{\mathrm{3}}]{\mathrm{7}}=\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$${or}\:\sqrt[{\mathrm{3}}]{\mathrm{7}}={x}^{\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${or}\:\sqrt[{\mathrm{3}}]{\mathrm{7}}={x}^{\left(\frac{\mathrm{3}−\mathrm{2}}{\mathrm{6}}\right)} \\ $$$${or}\:\sqrt[{\mathrm{3}}]{\mathrm{7}}={x}^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$${or}\:\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$${or}\:\mathrm{7}^{\frac{\mathrm{6}}{\mathrm{3}}} ={x}^{\frac{\mathrm{6}}{\mathrm{6}}} \\ $$$${or}\:\mathrm{7}^{\mathrm{2}} ={x}^{\mathrm{1}} \\ $$$$\boldsymbol{{So}}\:\underline{\underbrace{\boldsymbol{{x}}=\mathrm{49}}} \\ $$

Commented by Frix last updated on 08/Jun/25

You lost the 2^(nd) solution x=0 (7x)^(1/3) =x^(1/2) x^(1/2) −7^(1/3) x^(1/3) =0 x^(1/3) (x^(1/6) −7^(1/3) )=0 ⇒ x^(1/3) =0∨(x^(1/6) −7^(1/3) )=0 ⇒ x=0∨x=49

$$\mathrm{You}\:\mathrm{lost}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{solution}\:{x}=\mathrm{0} \\ $$$$\left(\mathrm{7}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{3}}} {x}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{0} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({x}^{\frac{\mathrm{1}}{\mathrm{6}}} −\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{0}\vee\left({x}^{\frac{\mathrm{1}}{\mathrm{6}}} −\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\vee{x}=\mathrm{49} \\ $$

Commented by fantastic last updated on 08/Jun/25