Question Number 34499 by mondodotto@gmail.com last updated on 07/May/18
$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\boldsymbol{\mathrm{cosh}}\left(\boldsymbol{\mathrm{ln}{x}}\right)−\boldsymbol{\mathrm{sinh}}\left(\boldsymbol{\mathrm{ln}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)=\mathrm{1}\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Answered by MJS last updated on 07/May/18
$$\mathrm{cosh}\:{x}=\frac{\mathrm{e}^{{x}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2e}^{{x}} } \\ $$$$\mathrm{sinh}\:{x}=\frac{\mathrm{e}^{{x}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2e}^{{x}} } \\ $$$$\mathrm{cosh}\:\mathrm{ln}\:{x}=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{x}} \\ $$$$\mathrm{sinh}\:\mathrm{ln}\:\frac{{x}}{\mathrm{2}}=\frac{{x}}{\mathrm{4}}−\frac{\mathrm{1}}{{x}} \\ $$$$\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{x}}−\left(\frac{{x}}{\mathrm{4}}−\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{7}}{\mathrm{4}} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{6}}{\mathrm{4}{x}}=\frac{\mathrm{7}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{6}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\mathrm{6} \\ $$