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} $

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}) $

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,

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

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

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

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