Question

The solution of the equation $2\cosh 2x+10\sinh 2x=5$ is

• A. $\frac{1}{2}\log \left(\frac{3}{5}\right)$
• B. $\frac{1}{2}\log \left(\frac{4}{3}\right)$
• C. $\frac{1}{2}\log \left(\frac{5}{4}\right)$
• D. $\frac{1}{2}\log \left(\frac{5}{3}\right)$

Answer 1

Answer: (B)

We have,
$2\cos h2x+10\sin h2x=5$
$2\left(\frac{e^{2x}+e^{-2x}}{2}\right)+10\left(\frac{e^{2x}-e^{-2x}}{2}\right)=5$
$2e^{2x}+2e^{-2x}+10e^{2x}-10e^{-2x}=10$
$12e^{2x}-8e^{-2x}-10=0$
$6e^{2x}-4^{-2x}-5=0$
$6{\left(e^{2x}\right)}^2-5e^{2x}-4=0$
$6{\left(e^{2x}\right)}^2-8e^{2x}+3e^{2x}-4=0$
$\left(2e^{2x}+1\right)\left(3e^{2x}-4\right)=0$
$\Rightarrow 3e^{2x}-4=0,2e^{2x}+1\ne 0$
$\Rightarrow e^{2x}=\frac{4}{3}$
$\Rightarrow 2x=\log \frac{4}{3}$
$\Rightarrow x=\frac{1}{2}\log \frac{4}{3}$