Logarithm

$\log _{(x+1)}\left(2 x^2+5 x+3\right)=4-\log _{(2 x+3)}\left(x^2+2 x+1\right)$

$\Rightarrow $ $\log _{(x+1)}(x+1)(2 x+3)=4-\log _{(2 x+3)}(x+1)^2$

$\Rightarrow $ $\log _{(x+1)}(x+1)+\log _{(x+1)}(2 x+3)=4-2 \log _{(2 x+3)}(x+1)$

$\Rightarrow $ $1+\frac{1}{\log _{(2 x+3)}(x+1)}=4-2 \log _{(2 x+3)}(x+1)$

Substitute $\log _{(2 x+3)}(x+1)=\boldsymbol{t}$

$\therefore $ $1+\frac{1}{\boldsymbol{t}}=4-2 \boldsymbol{t} \Rightarrow 2 \boldsymbol{t}^2-3 \boldsymbol{t}+1=0$

$\Rightarrow(\boldsymbol{t}-1)(2 \boldsymbol{t}-1)=0 \Rightarrow \boldsymbol{t}=\left\{1, \frac{1}{2}\right\}$

For t = 1,

$\log _{(2 x+3)}(x+1)=1 \Rightarrow 2 x+3=x+1 \Rightarrow x=-2$

For t = $\frac{1}{2}$,

$\log _{(2 x+3)}(x+1)=\frac{1}{2} \Rightarrow(x+1)^2=2 x+3 \Rightarrow x^2=2 \Rightarrow x= \pm \sqrt{2}$

j	xj	x + 1	Base	2x + 3	Base	2x2 + 5x + 3	Argument	x2 + 2x + 1	Argument	Remarks	Verdict
1	−2	−1	< 0	−1	< 0	1	> 0	1	> 0	1st and 2nd base −ve	Invalid
2	−√2	1 − √2	< 0	3 − 2√2	> 0	7 − 5√2	< 0	3 − 2√2	> 0	1st base and 1st argument −ve	Invalid
3	√2	1 + √2	> 0	3 + 2√2	> 0	7 + 5√2	> 0	3 + 2√2	> 0	All bases and arguments +ve	Valid

$\sum\limits_{j=1}^3\left[x_j\right]^2=0+0+(\sqrt{2})^2=2$

Given, $(2 a)^{\ln a}=(b c)^{\ln b}$, where $2 a>0, b c>0$

$ \Rightarrow \ln a(\ln 2+\ln a)=\ln b(\ln b+\ln c) $ ..........(i)

and $(b)^{\ln 2}=(a)^{\ln c}$

$ \Rightarrow \ln 2 \cdot \ln b=\ln c \cdot \ln a $ ..........(ii)

Now, let $\ln a=x, \ln b=y$

$ \ln 2=p, \ln c=z $

Now, from Eqs. (i) and (ii), we get

$ p \cdot y=x z \text { and } x(p+x)=y(y+z) $

$ \begin{aligned} & \therefore p=\frac{x z}{y} \\\\ & \therefore x\left(\frac{x z}{y}+x\right)=y(y+z) \\\\ & \Rightarrow x^2 z+x^2 y=y^2(y+z) \\\\ & \Rightarrow \left(x^2-y^2\right)(y+z)=0 \end{aligned} $

$ \therefore $ $ x^2 = y^2 $

$ x = \pm y $

So, from the equations, there are two cases :

Case 1 :

$ x = y $

In this case, since x and y are natural logarithms of positive numbers a and b respectively, this implies that a = b. However, this cannot be true as a, b, and c are given to be distinct positive real numbers.

Case 2 :

x = -y

In this case, $ \ln a = -\ln b $

$ \Rightarrow a \times b = 1$

$ \Rightarrow b = \frac{1}{a} $

Also, y + z = 0 (from the equations above)

$ \Rightarrow \ln b + \ln c = 0 $

$ \Rightarrow \ln (b \times c) = 0 $

$ \Rightarrow b \times c = 1 $

Given $ b = \frac{1}{a} $ and $ b \times c = 1 $ $ \Rightarrow c = a $

Thus, in the case where x = -y, the possible values are :

$b = \frac{1}{a} $

$c = a$

$ \begin{aligned} & \text { If } b c=1 \Rightarrow(2 a)^{\ln a}=1 \\\\ & \Rightarrow a=1 / 2 \\\\ & \text { So, } 6 a+5 b c=6\left(\frac{1}{2}\right)+5=3+5=8 \end{aligned} $

$ \begin{aligned} & \log _2\left(9^{2 \alpha-4}+13\right)-\log _2\left(\frac{5}{2} \cdot 3^{2 \alpha-4}+1\right)=2 \\\\ & \Rightarrow \frac{9^{2 \alpha-4}+13}{\frac{5}{2} 3^{2 \alpha-4}+1}=4 \\\\ & \Rightarrow \alpha=2 \quad \text { or } \quad 3 \\\\ & \sum_{\alpha \in \mathrm{S}} \alpha=5 \text { and } \sum_{\alpha \in \mathrm{S}}(\alpha+1)^2=25 \\\\ & \Rightarrow x^2-50 x+25 \beta=0 \text { has real roots } \\\\ & \Rightarrow \beta \leq 25 \\\\ & \Rightarrow \beta_{\max }=25 \end{aligned} $

${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$

${\log _{(x + 1)}}(2x + 5)(x + 1) + 2{\log _{(2x + 5)}}(x + 1) = 4$

${\log _{(x + 1)}}(2x + 5) + 1 + 2{\log _{(2x + 5)}}(x + 1) = 4$

Put ${\log _{(x + 1)}}(2x + 5) = t$

$t + {2 \over t} = 3 \Rightarrow {t^2} - 3t + 2 = 0$

t = 1, 2

${\log _{(x + 1)}}(2x + 5) = 1$ & ${\log _{(x + 1)}}(2x + 5) = 2$

$x + 1 = 2x + 3$ & $2x + 5 = {(x + 1)^2}$

$x = - 4$ (rejected)

${x^2} = 4 \Rightarrow x = 2, - 2$ (rejected)

So, x = 2

No. of solution = 1

${\log _4}(x - 1) = {\log _2}(x - 3)$

$ \Rightarrow {1 \over 2}{\log _2}(x - 1) = {\log _2}(x - 3)$

$ \Rightarrow {\log _2}{(x - 1)^{1/2}} = {\log _2}(x - 3)$

$ \Rightarrow {(x - 1)^{1/2}} = {\log _2}(x - 3)$

$ \Rightarrow {(x - 1)^{1/2}} = x - 3$

$ \Rightarrow x - 1 = {x^2} + 9 - 6x$

$ \Rightarrow {x^2} - 7x + 10 = 0$

$ \Rightarrow (x - 2)(x - 5) = 0$

$ \Rightarrow x = 2,5$

But x $ \ne $ 2 because it is not satisfying the domain of given equation i.e. log₂(x $-$ 3) $ \to $ its domain x > 3

finally x is 5

$ \therefore $ No. of solutions = 1.

${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$

$ \Rightarrow $ ${\log _{{1 \over 2}}}\left| {\sin x} \right|$ + ${\log _{{1 \over 2}}}\left| {\cos x} \right|$ = 2

$ \Rightarrow $ ${\log _{{1 \over 2}}}\left( {\left| {\sin x\cos x} \right|} \right)$ = 2

$ \Rightarrow $ ${\left| {\sin x\cos x} \right| = {1 \over 4}}$

$ \Rightarrow $ sin 2x = $ \pm $ ${1 \over 2}$

$ \therefore $ Number of distinct solution = 8