Logarithms

$x^{\frac{3}{4}\left(\log _2 x\right)^2+\log _2 x-\frac{5}{4}}=\sqrt{2}$

Taking log both sides with base 2

$ \begin{aligned} & \Rightarrow \log _2 x^{\frac{3}{4}\left(\log _2 x\right)^2+\log _2 x-\frac{5}{4}}=\log _2 \sqrt{2} \\ & \Rightarrow\left[\frac{3}{4}\left(\log _2 x\right)^2+\log _2 x-\frac{5}{4}\right] \log _2 x=\frac{1}{2} \end{aligned} $

Let $\log _2 x=t$

$ \therefore \quad\left(\frac{3 t^2}{4}+t-\frac{5}{4}\right) t=\frac{1}{2} $

$ \begin{aligned} & \Rightarrow \quad 3 t^3+4 t^2-5 t-2=0 \\ & \Rightarrow \quad 3 t^2(t-1)+7 t(t-1)+2(t-1)=0 \\ & \Rightarrow \quad\left(3 t^2+7 t+2\right)(t-1)=0 \\ & \Rightarrow \quad(3 t+1)(t+2)(t-1)=0 \\ & \quad t=\frac{-1}{3},-21 \\ & \Rightarrow \quad \log _2 x=\frac{-1}{3},-21 \\ & \Rightarrow \quad x=2^{\frac{-1}{3}}, 2^{-2}, 2 \Rightarrow 3 \text { real solutions } \end{aligned} $

Since, $7-\log _3|x| \geq 0 \Rightarrow \log _3|x| \leq 7$

And $4-\log _3|x| \geq 0 \Rightarrow \log _3|x| \leq 4$

$ \Rightarrow|x| \leq 3^4=81 $

Now, $7-\log _3|x|=\left(4-\log _3|x|\right)^2$

$ \begin{aligned} & \Rightarrow 7-\log _3|x|=16-8 \log _3|x|+\left(\log _3|x|\right)^2=0 \\ & \Rightarrow\left(\log _3|x|\right)^2-7 \log _3|x|+9=0 \end{aligned} $

Let $t=\log _3|x|$, then

$ \begin{aligned} & t^2-7 t+9=0 \\ & \Rightarrow t=\frac{7 \pm \sqrt{49-36}}{2}=\frac{7 \pm \sqrt{13}}{2} \end{aligned} $

Since, $t=\log _3|x|$ must be a real number and we know that $0 \leq \log _3|x| \leq 4$

So, only roots satisfy the condition is

$ 0 \leq \frac{7-\sqrt{13}}{2} \leq 4 $

and this root provides two solutions for $x$.

$[\because|x|$ means both $x$ and $-x$ satisfying the equation $]$

Thus, the number of real solutions is 2 .

To find $\cosh(\log 4)$, we start by using the identity for the hyperbolic cosine function:

$ \cosh(x) = \frac{e^x + e^{-x}}{2} $

Substituting $x = \log 4$ into the formula, we have:

$ \cosh(\log 4) = \frac{e^{\log 4} + e^{-\log 4}}{2} $

We know that:

$ e^{\log 4} = 4 \quad \text{and} \quad e^{-\log 4} = \frac{1}{4} $

Substitute these values back into the equation:

$ \cosh(\log 4) = \frac{4 + \frac{1}{4}}{2} = \frac{\frac{16}{4} + \frac{1}{4}}{2} $

Simplify the expression:

$ = \frac{\frac{17}{4}}{2} = \frac{17}{8} $

Therefore, $\cosh(\log 4) = \frac{17}{8}$.

To solve this problem, let's use the relationship for the inverse hyperbolic cosine function:

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

Applying this to $\cosh^{-1} 2$, we get:

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

Simplifying under the square root:

$ \sqrt{2^2 - 1} = \sqrt{4 - 1} = \sqrt{3} $

So, the expression becomes:

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

$ \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 $

Given,

$\begin{aligned} & 4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2 x-1} \\ \Rightarrow & (2)^{2 x}+2^{2 x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}} \\ \Rightarrow & 2^{2 x}+\frac{2^{2 x}}{2}=3^x \cdot \sqrt{3}+\frac{3^x}{\sqrt{3}} \\ \Rightarrow & 2^{2 x}\left[1+\frac{1}{2}\right]=3^x\left[\sqrt{3}+\frac{1}{\sqrt{3}}\right] \\ \Rightarrow & 2^{2 x} \times \frac{3}{2}=3^x \times \frac{4}{\sqrt{3}} \\ \Rightarrow & 2^{2 x}(3 \sqrt{3})=3^x(8) \\ \Rightarrow & 2^{2 x}(3)^{\frac{3}{2}}=3^x \times(2)^3 \end{aligned}$

Comparing power of $2,2 x=3$

and comparing power of $3, x=\frac{3}{2}$

Both equations given us $x=\frac{3}{2}$.

$\begin{aligned} & \left\{x \in R / \frac{\sqrt{|x|^2-2|x|-8}}{\log \left(2-x-x^2\right)} \text { is a real number }\right\} \\ & \left\{x \in R / \frac{\sqrt{x^2-2 x-8}}{\log (2+x)(1-x)} \text { is a real number }\right\} \\ & =\left\{x \in R / \frac{\sqrt{(x-4)(x+2)}}{\log \{(2+x)(1-x)\}} \text { is a real number }\right\} \\ & =\phi \end{aligned}$

Given,

$\begin{aligned} & 4+6\left(e^{2 x}+1\right) \tanh x=11 \cosh x+11 \sinh x \\ & \Rightarrow \quad 4+6\left(e^{2 x}+1\right) \cdot \frac{\left(e^{2 x}-1\right)}{\left(e^{2 x}+1\right)} \\ & =11\left\{\frac{e^{2 x}+1}{2 e^x}+\frac{e^{2 x}-1}{2 e^x}\right\} \\ & =11\left\{\frac{2 e^{2 x}}{2 e^x}\right\}=11 e^x \\ & \Rightarrow \quad 4+6\left(e^{2 x}-1\right)=11 e^x \\ & \text { Let } e^x=P \\ & \Rightarrow \quad 4+6\left(P^2-1\right)=11 P \\ & \Rightarrow \quad 6 P^2-11 P-2=0 \\ & \Rightarrow \quad 6 P^2-12 P+P-2=0 \\ & \Rightarrow \quad(P-2)(6 P+1)=0 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad P=2 \text { and } \frac{-1}{6} \\ & \text { Thus, } e^x=2 \\ & \Rightarrow \quad x=\log _e 2 \end{aligned}$

$\begin{aligned} & 4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2 x-1} \\ & \Rightarrow \quad 4^x-3^x \cdot 3^{-\frac{1}{2}}=3^x \cdot 3^{\frac{1}{2}}-2^{2 x} \cdot 2^{-1} \\ & \Rightarrow \quad 4^x-3^x \cdot \frac{1}{\sqrt{3}}=\sqrt{3} \cdot 3^x-\frac{4^x}{2} \\ & \Rightarrow \quad 4^x+\frac{4^x}{2}=\sqrt{3} \cdot 3^x+\frac{1}{\sqrt{3}} \cdot 3^x \\ & \Rightarrow \quad 4^x\left(1+\frac{1}{2}\right)=3^x\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right) \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{3}{2} \cdot 4^x=\frac{4}{\sqrt{3}} \cdot 3^x \\ & \Rightarrow \quad \frac{4^x}{3^x}=\frac{2 \cdot 4}{3 \sqrt{3}} \\ & \Rightarrow \quad\left(\frac{4}{3}\right)^x=\frac{4^{\frac{1}{2}} \cdot 4^1}{3^1 \frac{1}{3^2}}=\left(\frac{4}{3}\right)^{\frac{1}{2}+1}=\left(\frac{4}{3}\right)^{\frac{3}{2}} \\ & \Rightarrow \quad x=\frac{3}{2} \end{aligned}$

${\log _{225}}70 = {{{{\log }_e}70} \over {{{\log }_e}225}} = {{{{\log }_e}7 + {{\log }_e}5 + {{\log }_e}2} \over {{{\log }_e}25 + {{\log }_e}9}}$

$ = {{1 + {{{{\log }_e}5} \over {{{\log }_e}7}} + {{{{\log }_e}2} \over {{{\log }_e}7}}} \over {2{{{{\log }_e}5} \over {{{\log }_e}7}} + 2{{{{\log }_e}3} \over {{{\log }_e}7}}}}$

$ = {{1 + {{\log }_7}5 + {{\log }_3}2 \times {{\log }_5}3 \times {{\log }_7}5} \over {2[{{\log }_7}5 + {{\log }_5}3 \times {{\log }_7}5]}}$

$ = {{1 + a + abc} \over {2[a + ab]}} = {{1 + a + abc} \over {2a(1 + b)}}$

$ \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}} $

We have,

${\log _5}{{(a + b)} \over 3} = {{{{\log }_5}a + {{\log }_5}b} \over 2}$

$ \Rightarrow 2{\log _5}\left( {{{a + b} \over 3}} \right) = {\log _5}ab$

$ \Rightarrow {\log _5}{\left( {{{a + b} \over 3}} \right)^2} = {\log _5}ab$

$ \Rightarrow {{{{(a + b)}^2}} \over 9} = ab$

$ \Rightarrow {(a + b)^2} = 9ab$

$ \Rightarrow {a^2} + {b^2} = 7ab$

$ \Rightarrow {a \over b} + {b \over a} = 7$

$ \Rightarrow {\left( {{a \over b} + {b \over a}} \right)^2} = 49$

$ \Rightarrow {{{a^2}} \over {{b^2}}} + {{{b^2}} \over {{a^2}}} + 2 = 49$

$ \Rightarrow {{{a^4} + {b^4}} \over {{a^2}{b^2}}} = 49 - 2 = 47$