Question Number 176310 by Linton last updated on 16/Sep/22
$${a}^{{lna}} ={b}^{{lnb}} \\ $$$${a}−{b}=\mathrm{1} \\ $$$${solve}\:{for}\:{a}\:{and}\:{b} \\ $$
Answered by mr W last updated on 16/Sep/22
$$\left(\mathrm{ln}\:{a}\right)^{\mathrm{2}} =\left(\mathrm{ln}\:{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{ln}\:{a}=\pm\mathrm{ln}\:{b} \\ $$$$\Rightarrow{a}={b}\:\left({rejected}\:{due}\:{to}\:{a}−{b}=\mathrm{1}\right) \\ $$$$\:\:\:\:{or}\:{a}=\frac{\mathrm{1}}{{b}} \\ $$$$\frac{\mathrm{1}}{{b}}−{b}=\mathrm{1} \\ $$$${b}^{\mathrm{2}} +{b}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}>\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{{b}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$