tanh^{-1}(frac{1}{2}) + operatorname{coth}^{- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We use the logarithmic definitions of inverse hyperbolic functions: $	anh^{-1}(x) = \frac{1}{2} \ln(\frac{1+x}{1-x})$ and $\operatorname{coth}^{-

Vedclass · Accepted Answer

$\log \sqrt{6}$

Vedclass · Answer

(A) We use the logarithmic definitions of inverse hyperbolic functions: $	anh^{-1}(x) = \frac{1}{2} \ln(\frac{1+x}{1-x})$ and $\operatorname{coth}^{-1}(x) = \frac{1}{2} \ln(\frac{x+1}{x-1})$.For $	anh^{-1}(\frac{1}{2})$:$	anh^{-1}(\frac{1}{2}) = \frac{1}{2} \ln(\frac{1+1/2}{1-1/2}) = \frac{1}{2} \ln(\frac{3/2}{1/2}) = \frac{1}{2} \ln(3) = \log \sqrt{3}$.For $\operatorname{coth}^{-1}(3)$:$\operatorname{coth}^{-1}(3) = \frac{1}{2} \ln(\frac{3+1}{3-1}) = \frac{1}{2} \ln(\frac{4}{2}) = \frac{1}{2} \ln(2) = \log \sqrt{2}$.Adding these results:$\log \sqrt{3} + \log \sqrt{2} = \log(\sqrt{3} \cdot \sqrt{2}) = \log \sqrt{6}$.

$\tanh^{-1}(\frac{1}{2}) + \operatorname{coth}^{-1}(3) = $

Similar Questions

If $x_n > x_{n-1} > \dots > x_2 > x_1 > 1$,then the value of $\log_{x_1} \log_{x_2} \log_{x_3} \dots \log_{x_n} (x_n^{x_{n-1}^{\dots^{x_1}}})$ is:

Let $x$ and $y$ be real numbers such that $x > 2y > 0$ and $2 \log (x - 2y) = \log x + \log y$. Then,the possible value$(s)$ of $\frac{x}{y}$ is/are:

The expression $\log_{a} x$ is defined for $(a > 0, a \neq 1)$ when:

The solution to the equation ${9^x} - {2^{x + 1/2}} = {2^{x + 3/2}} - {3^{2x - 1}}$ is:

The number of roots of the equation $\log(-2x) = 2\log(x+1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module