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. > …

Show that tanh⁻¹(dfrac12) = dfrac12loge 3

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

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The inverse hyperbolic tangent has the logarithmic form $\tanh^{-1}x = …

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Question 14 of 14

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

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

100% · 14/14 Questions

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

The inverse hyperbolic tangent has the logarithmic form $\tanh^{-1}x = …

←Previous questionQ5 — Prove that for any x ∈ R, sinh(3x) = 3sinh x + 4sinh³ x.

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

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

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

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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

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

Unlock the full solution

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

Unlock the full solution