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Question Number 170796 by Beginner last updated on 31/May/22

Answered by som(math1967) last updated on 31/May/22

$${\left[\frac{{\left(\sqrt{2}\right)}^{\sqrt{3}}\{{\left(\sqrt{3}\right)}^{\sqrt{2}}+{\left(\sqrt{2}\right)}^{\sqrt{3}}\}}{{\left(\sqrt{2}\right)}^{\sqrt{2}}\{{\left(\sqrt{3}\right)}^{\sqrt{2}}+{\left(\sqrt{2}\right)}^{\sqrt{3}}\}}\right]}^{\frac{1}{\sqrt{3}-\sqrt{2}}}$$$${\left(\sqrt{2}\right)}^{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}}=\sqrt{2}ans$$$$$$

Answered by floor(10²Eta[1]) last updated on 31/May/22

$$\sqrt{3}=\mathrm{b},\sqrt{2}=\mathrm{a}$$$${\left[\frac{{\mathrm{a}}^{\mathrm{b}}{\mathrm{b}}^{\mathrm{a}}+{\mathrm{a}}^{2\mathrm{b}}}{{\mathrm{a}}^{\mathrm{a}}{\mathrm{b}}^{\mathrm{a}}+{\mathrm{a}}^{\mathrm{b}}{\mathrm{a}}^{\mathrm{a}}}\right]}^{\frac{1}{\mathrm{b}-\mathrm{a}}}={\left[\frac{{\mathrm{a}}^{\mathrm{b}}({\mathrm{b}}^{\mathrm{a}}+{\mathrm{a}}^{\mathrm{b}})}{{\mathrm{a}}^{\mathrm{a}}({\mathrm{b}}^{\mathrm{a}}+{\mathrm{a}}^{\mathrm{b}})}\right]}^{\frac{1}{\mathrm{b}-\mathrm{a}}}={\left({\mathrm{a}}^{\mathrm{b}-\mathrm{a}}\right)}^{\frac{1}{\mathrm{b}-\mathrm{a}}}=\mathrm{a}=\sqrt{2}$$

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