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

> > > [!TLDR] > The inverse hyperbolic tangent has the logarithmic form $\tanh^{-1}x = \dfrac12\log_e\dfrac{1+x}{1-x}$; substituting $x=\dfrac12$ gives the required value directly. > > **Concept.** For $|x| > *(Derivation: if $y = \tanh^{-1}x$, then $x = \tanh y = \dfrac{e^y-e^{-y}}{e^y+e^{-y}}$. Solving for $e^{2y}$ gives $e^{2y} = \dfrac{1+x}{1-x}$, so $2y = \log_e\dfrac{1+x}{1-x}$.)* > > **Step 1.** Substitute $x = \dfrac12$: …

Show that tanh⁻¹((1)/(2)) = (1)/(2)loge 3

Question 14 of 16

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

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2024Subjective· 2mImportance★★★★★

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Concept understanding — Logarithmic Forms of Inverse Hyperbolic Functions

Every inverse hyperbolic function can be rewritten as an explicit natural logarithm, and this is what makes the inverse functions computable rather than merely abstract symbols. The three principal theorems are:

\sinh^{-1}x=\log\!\left(x+\sqrt{x^2+1}\right)\ \text{for all real }x,\qquad \cosh^{-1}x=\log\!\left(x+\sqrt{x^2-1}\right)\ \text{for }x\ge1,

\tanh^{-1}x=\frac12\log\frac{1+x}{1-x}\ \text{for }-1<x<1.

The proof strategy is identical in every case: set $y$ equal to the inverse function, write $x$ in terms of $e^y$ and $e^{-y}$ using the original definition, and multiply through by $e^y$ (or $e^{2y}$ ) to turn the equation into a genuine quadratic in $e^y$ . Solving that quadratic, exactly one root is admissible (the other is rejected because $e^y$ must be positive, or because it lies outside the principal branch), and taking a logarithm of the surviving root recovers $y$ . …

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Q.Show that tanh⁡−1(12)=12log⁡e3\tanh^{-1}\left(\dfrac{1}{2}\right) = \dfrac{1}{2}\log_e 3tanh−1(21​)=21​loge​3

Concept understanding — Logarithmic Forms of Inverse Hyperbolic Functions

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Q.Show that $\tanh^{-1}\left(\dfrac{1}{2}\right) = \dfrac{1}{2}\log_e 3$

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