Logarithms

$ \begin{aligned} &\text { Given, sinh } \log (3+\sqrt{8}) \text { ) }\\ &\begin{aligned} & =\frac{e^{\log (3+\sqrt{8})}-e^{-\log (3+\sqrt{8})}}{2} \\ & =\frac{(3+\sqrt{8})-\left(\frac{1}{3+\sqrt{8}}\right)}{2} \\ & =\frac{(3+\sqrt{8})-(3-\sqrt{8})}{2}=\frac{2 \sqrt{8}}{2}=\sqrt{8}=2^{\frac{3}{2}} \end{aligned} \end{aligned} $

To determine the range of the function $ f(x) = \log_{0.5}(x^4 - 2x^2 + 3) $, we start by setting $ x^2 = t $. The expression then becomes:

$ t^2 - 2t + 3 = y $

To analyze the quadratic $ t^2 - 2t + 3 $, we calculate the discriminant $ D $:

$ D = b^2 - 4ac = (-2)^2 - 4 \times 1 \times 3 = -8 $

Since $ D < 0 $, the quadratic equation has no real roots; hence, $ y = t^2 - 2t + 3 $ is always positive.

Next, we find the value of $ t $ that minimizes the quadratic:

$ t_{\min} = -\frac{b}{2a} = \frac{-(-2)}{2 \times 1} = 1 $

Substituting back to find the minimum value of $ y $:

$ y_{\min} = -\frac{D}{4a} = \frac{-(-8)}{4 \times 1} = 2 $

Given that the base of the logarithm, $ 0.5 $, is less than 1, the logarithmic function will be decreasing. Therefore, we can solve for when $ f(x) = \log_{0.5}(2) $:

$ \begin{align*} f(x) &= \log_{0.5}(2) \\ &= \log_{\left(\frac{1}{2}\right)}(2) \\ &= \log_{2^{-1}}(2) = -1 \cdot \log_2(2) = -1 \end{align*} $

Thus, the range of $ f(x) $ is $ (-\infty, -1] $.

To solve for $ x $ given that $\sinh (\log x) = -2$, we start with the definition of the hyperbolic sine function:

$ \sinh y = \frac{e^y - e^{-y}}{2} $

Substituting $\log x$ into the equation:

$ \sinh (\log x) = \frac{e^{\log x} - e^{-\log x}}{2} $

Using the property $e^{\log x} = x$, this becomes:

$ -2 = \frac{x - \frac{1}{x}}{2} $

Multiplying through by 2 to clear the fraction:

$ -4 = x - \frac{1}{x} $

Rearranging terms and multiplying by $x$ to eliminate the fraction:

$ x^2 + 4x - 1 = 0 $

This is a quadratic equation, which we solve using the quadratic formula:

$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

For our equation, $a = 1$, $b = 4$, and $c = -1$. Plugging in these values gives:

$ x = \frac{-4 \pm \sqrt{16 + 4}}{2} $

Simplifying further:

$ x = \frac{-4 \pm \sqrt{20}}{2} $

$ x = \frac{-4 \pm 2\sqrt{5}}{2} $

$ x = -2 \pm \sqrt{5} $

This results in two possible solutions:

$x = \sqrt{5} - 2$

$x = -(2 + \sqrt{5})$

However, since $\log x$ implies $x > 0$, the valid solution is:

$ x = \sqrt{5} - 2 $

$ \log \left(9+3\sqrt{2}(2+\sqrt{5})+4\sqrt{5}\right) $

and match it with the correct option.

First, expand the middle term:

$ 3\sqrt{2}(2+\sqrt{5})=6\sqrt{2}+3\sqrt{10} $

So the expression inside the logarithm becomes

$ 9+4\sqrt{5}+6\sqrt{2}+3\sqrt{10} $

Now let us recall the standard NCERT forms:

1. For $\cosh^{-1} x$

If

$ y=\cosh^{-1} x $

then

$ y=\log\left(x+\sqrt{x^2-1}\right) $

So,

$ \cosh^{-1} 3=\log(3+\sqrt{8})=\log(3+2\sqrt{2}) $

2. For $\sinh^{-1} x$

If

$ y=\sinh^{-1} x $

then

$ y=\log\left(x+\sqrt{x^2+1}\right) $

So,

$ \sinh^{-1} 2=\log(2+\sqrt{5}) $

But in the options we have $\sinh^{-1} 3$, so let us check that too:

$ \sinh^{-1} 3=\log(3+\sqrt{10}) $

Now observe the given expression carefully:

$ 9+4\sqrt{5}+6\sqrt{2}+3\sqrt{10} $

Let us try factorising it as a product of two standard logarithmic forms.

Check:

$ (3+2\sqrt{2})(3+\sqrt{5}) $

Expanding:

$ =9+3\sqrt{5}+6\sqrt{2}+2\sqrt{10} $

This does not match.

Now check:

$ (3+2\sqrt{2})(2+\sqrt{5}) $

Expanding:

$ =6+3\sqrt{5}+4\sqrt{2}+2\sqrt{10} $

Not a match.

Now check:

$ (3+2\sqrt{2})(3+\sqrt{10}) $

Expanding:

$ =9+3\sqrt{10}+6\sqrt{2}+2\sqrt{20} $

Since $\sqrt{20}=2\sqrt{5}$,

$ 2\sqrt{20}=4\sqrt{5} $

Hence

$ (3+2\sqrt{2})(3+\sqrt{10}) =9+3\sqrt{10}+6\sqrt{2}+4\sqrt{5} $

This exactly matches the given expression.

Therefore,

$ \log \left(9+3\sqrt{2}(2+\sqrt{5})+4\sqrt{5}\right) = \log\left[(3+2\sqrt{2})(3+\sqrt{10})\right] $

Using $\log(ab)=\log a+\log b$,

$ =\log(3+2\sqrt{2})+\log(3+\sqrt{10}) $

Now,

$ \log(3+2\sqrt{2})=\cosh^{-1} 3 $

and

$ \log(3+\sqrt{10})=\sinh^{-1} 3 $

So,

$ \log \left(9+3\sqrt{2}(2+\sqrt{5})+4\sqrt{5}\right) = \cosh^{-1} 3+\sinh^{-1} 3 $

Hence the correct option is

$ \boxed{\text{Option B}} $