Logarithm

$\begin{aligned} & \log _{(x+3)}\left(6 x^2+28 x+30\right)=5-2 \log _{(6 x+10)}\left(x^2+6 x+9\right)-1 \\ & \log _{(x+3)} 2\left(3 x^2+14 x+15\right)+2 \log _{(6 x+10)}\left(x^2+6 x+9\right)=5 \\ & \log _{(x+3)} 2(x+3)(3 x+5)+2 \log _{(6 x+10)}(x+3)^2=5 \\ & \log _{(x+3)}(x+3)+\log _{(x+3)}(6 x+10)+4 \log _{(6 x+10)}(x+3)=5 \\ & \log _{(x+3)}(6 x+10)+4 \log _{(6 x+10)}(x+3)=4 \quad\left\{\log _{(x+3)}(x+3)=1\right\}\end{aligned}$

Let, $\log _{(x+3)}(6 x+10)=t$

$\Rightarrow \log _{(6 x+10)}(x+3)=\frac{1}{t}$

$\therefore $ $t+4 \times \frac{1}{t}=4$

$\begin{aligned} & t^2-4 t+4=0 \\ & (t-2)^2=0 \Longrightarrow t=2\end{aligned}$

$\begin{aligned} & \text { Checking } x=1 \text { in equation (1) } \\ & \log _{(1+3)}\left(6(1)^2+28(1)+30\right)=5-2 \log _{(6(1)+10)}\left(1^2+6(1)+9\right) \\ & \log _4 64=5-2 \log _{16} 16 \\ & 3=5-2=3 \text { true } \\ & \text { Checking } x=-1 \text { in equation (1) } \\ & \log _{(-1+3)}\left(6(-1)^2+28(-1)+30\right)=5-2 \log _{(6(-1)+10)}\left((-1)^2+6(-1)+9\right) \\ & \log _2 8=5-2 \log _4 4 \\ & 3=3 \text { true } \\ & \text { So, solutions are } x=1 \text { and } x=-1 \\ & \text { Sum of solution }=1+(-1)=0\end{aligned}$

We first find the values of $\alpha$ and $\beta$.

Given

$ \alpha=\frac14+\frac18+\frac1{16}+\cdots $

This is an infinite geometric progression with

$ a=\frac14,\quad r=\frac12 $

So,

$ \alpha=\frac{a}{1-r}=\frac{\frac14}{1-\frac12}=\frac{\frac14}{\frac12}=\frac12 $

Now,

$ \beta=\frac13+\frac19+\frac1{27}+\cdots $

This is also an infinite geometric progression with

$ a=\frac13,\quad r=\frac13 $

Thus,

$ \beta=\frac{a}{1-r}=\frac{\frac13}{1-\frac13}=\frac{\frac13}{\frac23}=\frac12 $

So both are equal to $\frac12$.

Now the expression becomes

$ (0.2)^{\log_{\sqrt5}\left(\frac12\right)}+(0.04)^{\log_5\left(\frac12\right)} $

Convert decimals into fractions:

$ 0.2=\frac15,\qquad 0.04=\frac1{25}=5^{-2} $

So,

$ \left(\frac15\right)^{\log_{\sqrt5}\left(\frac12\right)}+\left(\frac1{25}\right)^{\log_5\left(\frac12\right)} $

That is,

$ 5^{-\log_{\sqrt5}\left(\frac12\right)}+5^{-2\log_5\left(\frac12\right)} $

Now use the change of base idea:

$ \log_{\sqrt5}\left(\frac12\right)=\frac{\log_5\left(\frac12\right)}{\log_5(\sqrt5)} $

Since

$ \log_5(\sqrt5)=\log_5\left(5^{1/2}\right)=\frac12 $

therefore

$ \log_{\sqrt5}\left(\frac12\right)=\frac{\log_5\left(\frac12\right)}{1/2}=2\log_5\left(\frac12\right) $

Hence first term:

$ 5^{-\log_{\sqrt5}(1/2)}=5^{-2\log_5(1/2)} $

So both terms are same:

$ 5^{-2\log_5(1/2)}+5^{-2\log_5(1/2)}=2\cdot 5^{-2\log_5(1/2)} $

Now use the identity

$ a^{\log_a x}=x $

Then

$ 5^{-2\log_5(1/2)}=5^{\log_5(1/2)^{-2}} \text{ is not correct} $

So let us simplify properly:

$ -2\log_5\left(\frac12\right)=\log_5\left(\frac12\right)^{-2}? $

Instead, use

$ k\log_a b=\log_a(b^k) $

Thus,

$ -2\log_5\left(\frac12\right)=\log_5\left(\left(\frac12\right)^{-2}\right)=\log_5(4) $

Therefore,

$ 5^{-2\log_5(1/2)}=5^{\log_5 4}=4 $

So the total value is

$ 2\times 4=8 $

Hence, the required value is

$ \boxed{8} $

So the correct option is Option C.

We begin with the equation:

$ e^{5(\log_e x)^2 + 3} = x^8, \; x > 0. $

Equating the exponents, we have:

$ 5(\log_e x)^2 + 3 = \log_e x^8 = 8 \log_e x. $

Let $ t = \log_e x $. Substituting, the equation becomes:

$ 5t^2 + 3 = 8t. $

Rewriting this as a quadratic equation:

$ 5t^2 - 8t + 3 = 0. $

Factoring the quadratic:

$ 5t^2 - 5t - 3t + 3 = 0, $

$ (5t - 3)(t - 1) = 0. $

Thus, the solutions for $ t $ are:

$ t = 1 $ implies $ \log_e x = 1 $, giving $ x = e $.

$ t = \frac{3}{5} $ implies $ \log_e x = \frac{3}{5} $, giving $ x = e^{\frac{3}{5}} $.

The product of all solutions is:

$ e^1 \times e^{\frac{3}{5}} = e^{1 + \frac{3}{5}} = e^{\frac{8}{5}}. $

$\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$

Domain :

$ \begin{aligned} & x+\frac{7}{2}>0 \\\\ & x>\frac{-7}{2} \\\\ & x+\frac{7}{2} \neq 1 \\\\ & x \neq \frac{-5}{2} \\\\ & \frac{x-7}{2 x-3} \neq 0 \\\\ & x \neq 7 \\\\ & x \neq \frac{3}{2} \end{aligned} $

$ \text { Taking intersection : } x \in\left(\frac{-7}{2}, \infty\right)-\left\{-\frac{5}{2}, \frac{3}{2}, 7\right\} $

Now $\log _{\mathrm{a}} \mathrm{b} \geq 0$ if $\mathrm{a}>1$ and $\mathrm{b} \geq 1$

$ \begin{gathered} \text { Or } \\\\ a \in(0,1) \text { and } b \in(0,1) \end{gathered} $

Case I : $ x+\frac{7}{2}>1 \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \geq 1 $

$ \therefore $ $ x > -\frac{5}{2} $

and

$ \begin{aligned} & (2 x-3)^2-(x-7)^2 \leq 0 \\\\ & (2 x-3+x-7)(2 x-3-x+7) \leq 0 \\\\ & (3 x-10)(x+4) \leq 0 \end{aligned} $

$ x \in\left[-4, \frac{10}{3}\right] $

$ \text { Intersection : } \mathrm{x} \in\left(\frac{-5}{2}, \frac{10}{3}\right] $

Case II : $ x+\frac{7}{2} \in(0,1) \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \in(0,1) $

$ \begin{aligned} & \therefore 0 < x+\frac{7}{2}<1 \\\\ & -\frac{7}{2} < x < \frac{-5}{2} \end{aligned} $

and

$ \begin{aligned} & \left(\frac{x-7}{2 x-3}\right)^2<1 \\\\ & \Rightarrow (x-7)^2 < (2 x-3)^2 \end{aligned} $

$ \therefore $ $ x \in(-\infty,-4) \cup\left(\frac{10}{3}, \infty\right) $

No common values of $\mathrm{x}$.

Hence intersection with feasible region.

We get $x \in\left(\frac{-5}{2}, \frac{10}{3}\right]-\left\{\frac{3}{2}\right\}$

Integral value of $x$ are $\{-2,-1,0,1,2,3\}$

No. of integral values $=6$

${\log _{\cos x}}\cot x + 4{\log _{\sin x}}\tan x = 1$

$ \Rightarrow {\log _{\cos x}}\cot x - 4{\log _{\sin x}}\cot x = 1$

$ \Rightarrow 1 - {\log _{\cos x}}\sin x - 4 - 4{\log _{\sin x}}\cos x = 1$

Let ${\log _{\cos x}}\sin x = t$

$t + {4 \over t} = 4$

$ \Rightarrow t = 2$

$\sin x = {\cos ^2}x$

$ \Rightarrow \sin x = 1 - {\sin ^2}x$

$ \Rightarrow {\sin ^2}x + \sin {x^{ - 1}} = 0$

$ \Rightarrow \sin x = {{ - 1\, \pm \,\sqrt 5 } \over 2}$

as $x \in \left( {0,{\pi \over 2}} \right)$

$\sin x = {{\sqrt 5 - 1} \over 2}$

$x = {\sin ^{ - 1}}\left( {{{ - 1 + \sqrt 5 } \over 2}} \right)$

$ \Rightarrow \alpha = - 1,\beta = 5$

$\alpha + \beta = 4$