Question

show that tanh -1(1/2)=1/2loge3​

Answer 1

The hyperbolic tangent function is defined as,

$\longrightarrow\boxed{\tanh x=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}}$

Let us derive the definition for it's inverse.

Assume,

$\longrightarrow y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

Multiplying both numerator and denominator by $e^x,$ we get,

$\longrightarrow y=\dfrac{e^{2x}-1}{e^{2x}+1}$

$\longrightarrow ye^{2x}+y=e^{2x}-1$

$\longrightarrow 1+y=e^{2x}-ye^{2x}$

$\longrightarrow 1+y=e^{2x}(1-y)$

$\longrightarrow e^{2x}=\dfrac{1+y}{1-y}$

Taking antilog,

$\longrightarrow 2x=\log_e\left(\dfrac{1+y}{1-y}\right)$

$\longrightarrow x=\dfrac{1}{2}\log_e\left(\dfrac{1+y}{1-y}\right)$

Thus the inverse hyperbolic tangent function is defined as,

$\longrightarrow\boxed{\tanh^{-1}x=\dfrac{1}{2}\log_e\left(\dfrac{1+x}{1-x}\right)}$

Put $x=\dfrac{1}{2}.$ Then,

$\longrightarrow \tanh^{-1}\left(\dfrac{1}{2}\right)=\dfrac{1}{2}\log_e\left(\dfrac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)$

$\longrightarrow \tanh^{-1}\left(\dfrac{1}{2}\right)=\dfrac{1}{2}\log_e\left(\dfrac{\left(\frac{3}{2}\right)}{\left(\frac{1}{2}\right)}\right)$

$\longrightarrow\underline{\underline{\tanh^{-1}\left(\dfrac{1}{2}\right)=\dfrac{1}{2}\log_e3}}$