The value of the expression $\log \left( 3+2\sqrt{2} \right)=$
A. ${{\sinh }^{-1}}3$
B. ${{\cosh }^{-1}}3$
C. ${{\tanh }^{-1}}3$
D. ${{\coth }^{-1}}3$

Hint: In order to solve this problem, we must require the four inverse hyperbolic function formulae which are,
${{\sinh }^{-1}}x=\ln \left( x+\sqrt{{{x}^{2}}+1} \right)$
${{\cosh }^{-1}}x=\ln \left( x+\sqrt{{{x}^{2}}-1} \right)$
${{\tanh }^{-1}}x=\dfrac{1}{2}\ln \dfrac{1-x}{1+x}$
${{\coth }^{-1}}x=\dfrac{1}{2}\ln \dfrac{1+x}{1-x}$
We then write $\log \left( 3+2\sqrt{2} \right)$ after manipulating it as $\log \left( 3+\sqrt{{{3}^{2}}-1} \right)$ and then compare this form with the four formulae above, and then find the correct option

Complete step by step solution:
The basic inverse hyperbolic functions are ${{\sinh }^{-1}}x,{{\cosh }^{-1}}x,{{\tanh }^{-1}}x,{{\coth }^{-1}}x$ . The formulae for these inverse hyperbolic functions are,
${{\sinh }^{-1}}x=\ln \left( x+\sqrt{{{x}^{2}}+1} \right)$
${{\cosh }^{-1}}x=\ln \left( x+\sqrt{{{x}^{2}}-1} \right)$
${{\tanh }^{-1}}x=\dfrac{1}{2}\ln \dfrac{1-x}{1+x}$
${{\coth }^{-1}}x=\dfrac{1}{2}\ln \dfrac{1+x}{1-x}$
The given expression that we have in this problem is $\log \left( 3+2\sqrt{2} \right)$ . Now, we can write $2\sqrt{2}$ as $\sqrt{4\times 2}$ which is nothing but $\sqrt{8}$ . Now, we can write $\sqrt{8}$ as $\sqrt{9-1}$ . $9$ is nothing but the square of $3$ . So, $\sqrt{9-1}=\sqrt{{{3}^{2}}-1}$ . So, we can rewrite the entire expression $\log \left( 3+2\sqrt{2} \right)$ as,
$\Rightarrow \log \left( 3+2\sqrt{2} \right)=\log \left( 3+\sqrt{{{3}^{2}}-1} \right)$
We can find some similarity between the above expression and the four formulae that we have written before. If we compare $\log \left( 3+\sqrt{{{3}^{2}}-1} \right)$ and the formula for ${{\cosh }^{-1}}x$ , we can see that there is a lot common between the two forms of expression. If we put x as $3$ in the formula for ${{\cosh }^{-1}}x$ , we get,
${{\cosh }^{-1}}3=\ln \left( 3+\sqrt{{{3}^{2}}-1} \right)$
This is nothing but the expression that we had derived.
Thus, we can conclude that the value of $\log \left( 3+2\sqrt{2} \right)$ is ${{\cosh }^{-1}}3$ , which is option B.

Note: In order to solve these types of problems, we must remember the four basic inverse hyperbolic functions formulae. This helps us in solving the problems quickly, or we would have to derive the formula for them which would become tedious. We can also find the value of $\log \left( 3+2\sqrt{2} \right)$ using a calculator and compare the value with the values of ${{\sinh }^{-1}}3,{{\cosh }^{-1}}3,{{\tanh }^{-1}}3,{{\coth }^{-1}}3$ , found with a calculator.