Question

If $\tanh x=\frac{1}{4}$, then whether the given equation $x=\frac{1}{2}+{\log }_e\left(\frac{5}{3}\right)$ is ?

• A. True
• B. False

Answer 1

The correct option is A True

Given

$\tanh x=\frac{1}{4}$

$⟹\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{1}{4}$

$(\tanh x=\frac{e^{2x}-1}{e^{2x}+1})$

$⟹\frac{e^{2x}-1}{e^{2x}+1}=\frac{1}{4}$

$⟹4e^{2x}-4=e^{2x}+1$

$⟹3e^{2x}=5$

$⟹e^{2x}=\frac{5}{3}$

$⟹2x={\log }_e\left(\frac{5}{3}\right)$

$⟹x=\frac{1}{2}{\log }_e\left(\frac{5}{3}\right)$