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Question Number 188621 by 073 last updated on 03/Mar/23

Answered by HeferH last updated on 04/Mar/23

$$\frac{a}{b}=\frac{2}{3}=\frac{2k}{3k}$$$$a^b=b^a$$$${\left(2k\right)}^{3k}={\left(3k\right)}^{2k}$$$$2^{3k}\cdot k^{3k}=3^{2k}\cdot k^{2k}$$$${\left(2^3\right)}^k\cdot {\left(k^k\right)}^3={\left(3^2\right)}^k\cdot {\left(k^k\right)}^2$$$$8^k\cdot k^k=9^k$$$$k^k=\frac{9^k}{8^k}={\left(\frac{9}{8}\right)}^k\Rightarrow k=\frac{9}{8}$$$$b-a=3k-2k=k=\frac{9}{8}$$

Commented by 073 last updated on 03/Mar/23

$$\mathrm{nice}\mathrm{solution}$$

Answered by Frix last updated on 04/Mar/23

$$b=\frac{3a}{2}$$$$a^b=b^a$$$$a^{\frac{3a}{2}}={\left(\frac{3a}{2}\right)}^a\iff a^{\frac{3}{2}}=\frac{3a}{2}\iff a^{\frac{1}{2}}=\frac{3}{2}\Rightarrow a=\frac{9}{4}$$$$b-a=\frac{3a}{2}-a=\frac{a}{2}=\frac{9}{8}$$

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