Question Number 221322 by Frix last updated on 30/May/25
$$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\frac{\mathrm{1}}{{a}}} +\sqrt{{x}^{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}} }={x}^{\frac{\mathrm{1}}{{b}}} \\ $$
Answered by fantastic last updated on 30/May/25
![or x^(1/a) +(√x^((a+b)/(ab)) )=x^(1/b) or x^((a+b)/(2ab)) = x^(1/b) −x^(1/a) or x^((a+b)/(2ab)) =(x^(1/(2b)) )^2 −(x^(1/(2a)) )^2 or (x^(1/(2b)) )^2 −x^((1/(2a))+(1/(2b))) −(x^(1/(2a)) )^2 =0 or (x^(1/(2b)) )^2 −x^(1/(2b)) .x^(1/(2a)) −(x^(1/(2a)) )^2 =0 let x^(1/(2b)) be v and x^((1/(2a)) ) be u then v^2 −vu−u^2 =0 or v=((−(−u)±(√(u^2 −4.1.(−u^2 ))))/(2.1))=((u±(√(5u^2 )))/2)=((u±u(√5))/2)=u(((1±(√5))/2)) so v=u(((1±(√5))/2)) so v=u(((1+(√5))/2))[∵((1−(√5))/2)<0 , so rejected] but v=x^(1/(2b)) and u=x^(1/(2a)) so x^(1/(2b)) =x^(1/(2a)) (((1+(√5))/2)) or x^((((2ab)/(2b)))) =x^((((2ab)/(2a)))) (((1+(√5))/2))^(2ab) or x^a =x^b (((1+(√5))/2))^(2ab) or (x^a /x^b )=(((1+(√5))/2))^(2ab) or x^((a−b)) =(((1+(√5))/2))^(2ab) or x^(((a−b)×(1/((a−b))))) =(((1+(√5))/2))^((2ab)/((a−b))) or x^1 =(((1+(√5))/2))^((2ab)/((a−b))) So x=(((1+(√5))/2))^((2ab)/((a−b))) ✓✓](https://www.tinkutara.com/question/Q221325.png)
$${or}\:{x}^{\frac{\mathrm{1}}{{a}}} +\sqrt{{x}^{\frac{{a}+{b}}{{ab}}} }={x}^{\frac{\mathrm{1}}{{b}}} \: \\ $$$${or}\:{x}^{\frac{{a}+{b}}{\mathrm{2}{ab}}} =\:{x}^{\frac{\mathrm{1}}{{b}}} −{x}^{\frac{\mathrm{1}}{{a}}} \: \\ $$$${or}\:{x}^{\frac{{a}+{b}}{\mathrm{2}{ab}}} =\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}} \right)^{\mathrm{2}} −\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}} \right)^{\mathrm{2}} \\ $$$${or}\:\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}} \right)^{\mathrm{2}} −{x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}+\frac{\mathrm{1}}{\mathrm{2}{b}}} −\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${or}\:\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}} \right)^{\mathrm{2}} −{x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}} .{x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}} −\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:{x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}} \:{be}\:{v}\:{and}\:{x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}\:} \:{be}\:{u}\:{then} \\ $$$$\:{v}^{\mathrm{2}} −{vu}−{u}^{\mathrm{2}} =\mathrm{0} \\ $$$${or}\:{v}=\frac{−\left(−{u}\right)\pm\sqrt{{u}^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\left(−{u}^{\mathrm{2}} \right)}}{\mathrm{2}.\mathrm{1}}=\frac{{u}\pm\sqrt{\mathrm{5}{u}^{\mathrm{2}} }}{\mathrm{2}}=\frac{{u}\pm{u}\sqrt{\mathrm{5}}}{\mathrm{2}}={u}\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$${so}\:{v}={u}\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:{so}\:{v}={u}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left[\because\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\:,\:{so}\:{rejected}\right] \\ $$$${but}\:{v}={x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}} \:{and}\:{u}={x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}} \:{so} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$${or}\:{x}^{\left(\frac{\mathrm{2}{ab}}{\mathrm{2}{b}}\right)} ={x}^{\left(\frac{\mathrm{2}{ab}}{\mathrm{2}{a}}\right)} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{ab}} \\ $$$${or}\:{x}^{{a}} ={x}^{{b}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{ab}} \\ $$$${or}\:\:\frac{{x}^{{a}} }{{x}^{{b}} }=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{ab}} \\ $$$${or}\:{x}^{\left({a}−{b}\right)} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{ab}} \\ $$$${or}\:{x}^{\left(\left({a}−{b}\right)×\frac{\mathrm{1}}{\left({a}−{b}\right)}\right)} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\frac{\mathrm{2}{ab}}{\left({a}−{b}\right)}} \\ $$$${or}\:{x}^{\mathrm{1}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\frac{\mathrm{2}{ab}}{\left({a}−{b}\right)}} \\ $$$$\mathrm{S}{o}\:{x}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\frac{\mathrm{2}{ab}}{\left({a}−{b}\right)}} \checkmark\checkmark \\ $$
Commented by Frix last updated on 30/May/25
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Answered by mr W last updated on 30/May/25
$${a}\neq{b},\:{otherwise}\:{no}\:{solution}. \\ $$$${say}\:{A}={x}^{\frac{\mathrm{1}}{{a}}} ,\:{B}={x}^{\frac{\mathrm{1}}{{b}}} \\ $$$${A}+\sqrt{{AB}}={B} \\ $$$$\frac{{A}}{{B}}+\sqrt{\frac{{A}}{{B}}}=\mathrm{1} \\ $$$${t}^{\mathrm{2}} +{t}−\mathrm{1}=\mathrm{0}\:\:\:{with}\:{t}=\sqrt{\frac{{A}}{{B}}}\:>\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\:\left(\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\:{rejected}\right) \\ $$$$\frac{{A}}{{B}}={t}^{\mathrm{2}} =\mathrm{1}−{t}=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\frac{{A}}{{B}}={x}^{\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}} ={x}^{\frac{{b}−{a}}{{ab}}} \\ $$$$\Rightarrow{x}=\left(\frac{{A}}{{B}}\right)^{\frac{{ab}}{{b}−{a}}} =\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\frac{{ab}}{{b}−{a}}} \:\checkmark \\ $$
Commented by fantastic last updated on 30/May/25
$${Is}\:{not}\:\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\frac{{ab}}{{b}−{a}}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\frac{\mathrm{2}{ab}}{{a}−{b}}} ?? \\ $$
Commented by mr W last updated on 30/May/25
$${yes}. \\ $$
Commented by fantastic last updated on 30/May/25
yay!!
Commented by Frix last updated on 30/May/25
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Answered by Frix last updated on 30/May/25
![x^(1/b) −x^(1/a) =x^((1/(2a))+(1/(2b))) x^(1/a) (x^((1/b)−(1/a)) −1)=x^(1/a) x^((1/(2b))−(1/(2a))) x_1 =0 x^((1/b)−(1/a)) −x^((1/(2b))−(1/(2a))) −1=0 (x^((1/(2b))−(1/(2a))) )^2 −x^((1/(2b))−(1/(2a))) −1=0 x^((1/(2b))−(1/(2a))) =(1/2)±((√5)/2) x_(2, 3) =((1/2)±((√5)/2))^((2ab)/(a−b)) [Not all solutions exist depending on the values of a and b]](https://www.tinkutara.com/question/Q221334.png)
$${x}^{\frac{\mathrm{1}}{{b}}} −{x}^{\frac{\mathrm{1}}{{a}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}{a}}+\frac{\mathrm{1}}{\mathrm{2}{b}}} \\ $$$${x}^{\frac{\mathrm{1}}{{a}}} \left({x}^{\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}} −\mathrm{1}\right)={x}^{\frac{\mathrm{1}}{{a}}} {x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}−\frac{\mathrm{1}}{\mathrm{2}{a}}} \\ $$$${x}_{\mathrm{1}} =\mathrm{0} \\ $$$${x}^{\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}} −{x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}−\frac{\mathrm{1}}{\mathrm{2}{a}}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}−\frac{\mathrm{1}}{\mathrm{2}{a}}} \right)^{\mathrm{2}} −{x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}−\frac{\mathrm{1}}{\mathrm{2}{a}}} −\mathrm{1}=\mathrm{0} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{2}{b}}−\frac{\mathrm{1}}{\mathrm{2}{a}}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\frac{\mathrm{2}{ab}}{{a}−{b}}} \\ $$$$\left[\mathrm{Not}\:\mathrm{all}\:\mathrm{solutions}\:\mathrm{exist}\:\mathrm{depending}\:\mathrm{on}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{values}\:\mathrm{of}\:{a}\:\mathrm{and}\:{b}\right] \\ $$