Question Number 222339 by Tawa11 last updated on 22/Jun/25
$$\mathrm{Solve}:\:\:\:\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{36}}\:\:\:+\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{24}}\:\:\:\:=\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{16}} \\ $$
Answered by fantastic last updated on 22/Jun/25
![(36)^(1/x) +(24)^(1/x) =(16)^(1/x) or (((36)/(16)))^(1/x) +(((24)/(16)))^(1/x) =1 or ((9/4))^(1/x) +((3/2))^(1/x) =1 or ((3/2))^(2/x) +((3/2))^(1/x) =1 or y^2 +y=1[y=((3/2))^(1/x) ] y=((−1±(√5))/2) So y=(((√5)−1)/2)[∵((−1−(√5))/2)<0] ∴((3/2))^(1/x) =((((√5)−1)/2)) ∴x=log _(((√5)−1)/2) ((3/2))≈−0.84259173817672](https://www.tinkutara.com/question/Q222349.png)
$$\left(\mathrm{36}\right)^{\frac{\mathrm{1}}{{x}}} +\left(\mathrm{24}\right)^{\frac{\mathrm{1}}{{x}}} =\left(\mathrm{16}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$${or}\:\left(\frac{\mathrm{36}}{\mathrm{16}}\right)^{\frac{\mathrm{1}}{{x}}} +\left(\frac{\mathrm{24}}{\mathrm{16}}\right)^{\frac{\mathrm{1}}{{x}}} =\mathrm{1} \\ $$$${or}\:\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{\frac{\mathrm{1}}{{x}}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} =\mathrm{1} \\ $$$${or}\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{{x}}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} =\mathrm{1} \\ $$$${or}\:{y}^{\mathrm{2}} +{y}=\mathrm{1}\left[{y}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} \right] \\ $$$${y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${So}\:{y}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\left[\because\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\right] \\ $$$$\therefore\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} =\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\therefore{x}=\mathrm{log}\:_{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)\approx−\mathrm{0}.\mathrm{84259173817672} \\ $$
Commented by Tawa11 last updated on 22/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by Ghisom last updated on 22/Jun/25
$$\mathrm{36}^{\mathrm{1}/{x}} +\mathrm{24}^{\mathrm{1}/{x}} =\mathrm{16}^{\mathrm{1}/{x}} \\ $$$$\left(\mathrm{2}^{\mathrm{2}} \mathrm{3}^{\mathrm{2}} \right)^{\mathrm{1}/{x}} +\left(\mathrm{2}^{\mathrm{3}} \mathrm{3}\right)^{\mathrm{1}/{x}} =\left(\mathrm{2}^{\mathrm{4}} \right)^{\mathrm{1}/{x}} \\ $$$$\left(\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{1}/{x}} +\left(\mathrm{2}×\mathrm{3}\right)^{\mathrm{1}/{x}} =\left(\mathrm{2}^{\mathrm{2}} \right)^{\mathrm{1}/{x}} \\ $$$${t}=\left(\mathrm{3}/\mathrm{2}\right)^{\mathrm{1}/{x}} \wedge{t}>\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} +{t}\right){t}^{\frac{\mathrm{4ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:−\mathrm{ln}\:\mathrm{2}}} ={t}^{\frac{\mathrm{4ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:−\mathrm{ln}\:\mathrm{2}}} \\ $$$${t}^{\mathrm{2}} +{t}=\mathrm{1} \\ $$$${t}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{ln}\:\mathrm{3}\:−\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\:−\mathrm{ln}\:\mathrm{2}}\approx−.\mathrm{842591738} \\ $$
Commented by Tawa11 last updated on 22/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$