The sum of all the real solutions of the equation $\log _{(x+3)}\left(6 x^2+28 x+30\right)=5-2 \log _{(6 x+10)}\left(x^2+6 x+9\right)$ is equal to :
1
4
0
2
The product of all solutions of the equation $\mathrm{e}^{5\left(\log _{\mathrm{e}} x\right)^2+3}=x^8, x>0$, is :
The number of integral solutions $x$ of $\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$ is :
If the solution of the equation $\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1, x \in\left(0, \frac{\pi}{2}\right)$, is $\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)$, where $\alpha$, $\beta$ are integers, then $\alpha+\beta$ is equal to :
Let a, b, c be three distinct positive real numbers such that ${(2a)^{{{\log }_e}a}} = {(bc)^{{{\log }_e}b}}$ and ${b^{{{\log }_e}2}} = {a^{{{\log }_e}c}}$.
Then, 6a + 5bc is equal to ___________.
Explanation:
$ \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} $
Let $S = \left\{ {\alpha :{{\log }_2}({9^{2\alpha - 4}} + 13) - {{\log }_2}\left( {{5 \over 2}.\,{3^{2\alpha - 4}} + 1} \right) = 2} \right\}$. Then the maximum value of $\beta$ for which the equation ${x^2} - 2{\left( {\sum\limits_{\alpha \in s} \alpha } \right)^2}x + \sum\limits_{\alpha \in s} {{{(\alpha + 1)}^2}\beta = 0} $ has real roots, is ____________.
Explanation:
${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$, x > 0, is :
Explanation:
${\log _{(x + 1)}}(2x + 5)(x + 1) + 2{\log _{(2x + 5)}}(x + 1) = 4$
${\log _{(x + 1)}}(2x + 5) + 1 + 2{\log _{(2x + 5)}}(x + 1) = 4$
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
Explanation:
$ \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.
${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$ in the interval [0, 2$\pi $], is ____.
Explanation:
$ \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