Question

Solve the following equations:
(a) $\cosh \left(\ln x\right)-\sinh \left(\ln \frac{x}{2}\right)=1\frac{3}{4}$
(b) $4\sin hx+3e^x+3=0$

Answer 1

$a)\cosh \left(\ln x\right)-\sinh \left(\ln \frac{x}{2}\right)=1\frac{3}{4}$

$\Rightarrow \frac{e^{\ln x}-e^{-\ln x}}{2}-\frac{e^{\ln \frac{x}{2}}-e^{-\ln \frac{x}{2}}}{2}=\frac{7}{4}$

$\Rightarrow \frac{x+\frac{1}{x}}{2}-\frac{\frac{x}{2}-\frac{2}{x}}{2}=\frac{7}{4}$

$\Rightarrow \frac{x}{2}+\frac{1}{2x}-\frac{x}{4}+\frac{1}{x}=\frac{7}{4}$

$\Rightarrow \frac{2x-x}{4}+\frac{1}{x}(\frac{1}{2}+1)=\frac{7}{4}$

$\Rightarrow \frac{x}{4}+\frac{3}{2x}=\frac{7}{4}$

$\Rightarrow \frac{{2x}^2+12}{8x}=\frac{7}{4}$

$\Rightarrow 2x^2+12=14x$

$\Rightarrow x^2+6=7x\Rightarrow x^2-7x+6=0$

$\Rightarrow x^2-6x-x+6=0\Rightarrow x(x-6)-1(x-6)=0$

$\Rightarrow x=1,6$

$b)4\sinh x+3e^x+3=0$

$\because \sinh x=\frac{e^x-e^{-x}}{2}$

$\Rightarrow \frac{4(e^x-e^{-x})}{2}+3e^x+3=0$

$\Rightarrow 2e^x-2e^{-x}+3e^x+3=0$

$\Rightarrow 5e^x-2e^{-x}+3=0$

$\Rightarrow 5e^x-\frac{2}{e^x}+3=0$

$\Rightarrow 5e^{2x}-2+3e^x=0$

$\Rightarrow 5e^{2x}+3e^x-2=0$

$\Rightarrow 5e^{2x}+5e^x-2e^x-2=0$

$\Rightarrow 5e^x(e^x+1)-2{(e}^x+1)=0$

$\Rightarrow e^x=\frac{2}{5}$ or, $e^x=-1$ (not possible)

$\Rightarrow e^x=\frac{2}{5}$

$\Rightarrow \log e^x=\log \frac{2}{5}$

$\Rightarrow x={\log }_{e}2-{\log }_{e}5$