Question

Find the value of $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$, here logarithm is to the natural base e.
(a) 2
(b) 3
(c) $2-\sqrt{5}$
(d) $2+\sqrt{5}$

Answer 1

We start solving the problem by recalling the definition of sine hyperbolic function as $\sinh \left( y \right)=\dfrac{{{e}^{y}}-{{e}^{-y}}}{2}$. We then substitute $\log \left( 2+\sqrt{5} \right)$ in place of ‘y’ and then make use of the results $-\log a=\log \dfrac{1}{a}$ and ${{e}^{\log a}}=a$ to proceed through the problems. We then make use of the result ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and then perform the necessary calculations to get the required result.

Complete step-by-step answer:
According to the problem, we are asked to find the value of $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$, where logarithm is to the natural base e.
We know that the sine hyperbolic function is defined as $\sinh \left( y \right)=\dfrac{{{e}^{y}}-{{e}^{-y}}}{2}$. Let us use this definition to find the value of the given $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$.
So, we have $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{{{e}^{\log \left( 2+\sqrt{5} \right)}}-{{e}^{-\log \left( 2+\sqrt{5} \right)}}}{2}$ ---(1).
We know that $-\log a=\log \dfrac{1}{a}$. Let us use this result in equation (1).
$\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{{{e}^{\log \left( 2+\sqrt{5} \right)}}-{{e}^{\log \left( \dfrac{1}{2+\sqrt{5}} \right)}}}{2}$ ---(2).
We know that ${{e}^{\log a}}=a$. Let us use this result in equation (2).
$\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{2+\sqrt{5}-\left( \dfrac{1}{2+\sqrt{5}} \right)}{2}$.
$\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( \dfrac{{{\left( 2+\sqrt{5} \right)}^{2}}-1}{2+\sqrt{5}} \right)}{2}$ ---(3).
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Let us use this result in equation (3).
$\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( \dfrac{{{2}^{2}}+{{\left( \sqrt{5} \right)}^{2}}+2\left( 2 \right)\left( \sqrt{5} \right)-1}{2+\sqrt{5}} \right)}{2}$.
$\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( 4+5+4\sqrt{5} \right)-1}{2\left( 2+\sqrt{5} \right)}$.
$\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{8+4\sqrt{5}}{4+2\sqrt{5}}$.
$\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=2$.
So, we have found the value of $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$ as 2.
∴ The correct option for the given problem is (a).

So, the correct answer is “Option (a)”.

Note: If the given logarithm is not to the natural base, then the result will not be the same as we get in this problem. We should always take logarithms to the natural base if nothing is mentioned in the problem about the base while solving problems related to hyperbolic functions. We can also solve this problem as shown below:
Let us assume $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)=x$.
$\Rightarrow \log \left( 2+\sqrt{5} \right)={{\sinh }^{-1}}\left( x \right)$.
We know that ${{\sinh }^{-1}}\left( x \right)=\log \left( x+\sqrt{{{x}^{2}}+1} \right)$.
$\Rightarrow \log \left( 2+\sqrt{5} \right)=\log \left( x+\sqrt{{{x}^{2}}+1} \right)$.
$\Rightarrow 2+\sqrt{5}=x+\sqrt{{{x}^{2}}+1}$.
Comparing rational and irrational parts on both sides we get,
$\Rightarrow x=2$.