2 tanh^{-1} frac{1}{2} is equal to | vedclass

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(C) We use the formula $2 	anh^{-1} x = 	anh^{-1} \left( \frac{2x}{1+x^2} ight)$.Substituting $x = \frac{1}{2}$:$2 	anh^{-1} \left( \frac{1}{2} \

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$\log 3$

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(C) We use the formula $2 	anh^{-1} x = 	anh^{-1} \left( \frac{2x}{1+x^2} ight)$.Substituting $x = \frac{1}{2}$:$2 	anh^{-1} \left( \frac{1}{2} ight) = 	anh^{-1} \left( \frac{2(\frac{1}{2})}{1+(\frac{1}{2})^2} ight) = 	anh^{-1} \left( \frac{1}{1 + \frac{1}{4}} ight) = 	anh^{-1} \left( \frac{1}{\frac{5}{4}} ight) = 	anh^{-1} \left( \frac{4}{5} ight)$.Now, we use the logarithmic form of the inverse hyperbolic tangent function: $	anh^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} ight)$.Substituting $x = \frac{4}{5}$:$	anh^{-1} \left( \frac{4}{5} ight) = \frac{1}{2} \log \left( \frac{1 + \frac{4}{5}}{1 - \frac{4}{5}} ight) = \frac{1}{2} \log \left( \frac{\frac{9}{5}}{\frac{1}{5}} ight) = \frac{1}{2} \log 9 = \frac{1}{2} \log 3^2 = \log 3$.

$2 \tanh^{-1} \frac{1}{2}$ is equal to

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