$\because \displaystyle \log _{ n }{ m } =\frac { \log{ m }  }{ \log { n }  } $
$\therefore \log _an\times \log _nm=\dfrac{\log n}{\log a}\times \dfrac{\log m}{\log n}=\dfrac{\log m}{\log a}=\log _am$

 $ \log _{10}\left [ \log _{10}\left ( \log _{10}x \right ) \right ]=0 $
$log _{10} log _{10} x=1$
$log _{10 }x=10$
$x=10^{10}$

$\log _5 x =y $

$\log _{243}x=0.8$

$\log _x(7x-10) =2$

$\Rightarrow x^2 =7x-10$ ...... $[\because a^{\log _{a} x}=x]$

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

$\Rightarrow x^2-5x-2x-10 =0$

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

Hence, $x =5 \; or \; x =2$

Let $y = (\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 )$

Then, $2^y =2^{((\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

By laws of exponents and the definition of a logarithm,

$2^{((\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (log _{31} 32 ))} $

$=(2^{(\log _23)})^{((\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

$= 3^{((\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

$=(3^{(\log _34)})^{(( \log _4 5 ) … (\log _{31} 32 ))} $

$=4^{(( \log _4 5 ) … (\log _{31} 32 ))} $ .......

$=31^{(\log _{31}32)} =32$

$\therefore 2^y =32$

$y=5$

$\log _{49}7 = \log _{7^2}7 = \dfrac{1}{2}\log _{7}7 = \dfrac{1}{2}$

$\log _3(27)$

Since $\log _{ a }{ b } =\cfrac { \log _{ c }{ b }  }{ \log _{ c }{ a }  } $, we have with base $x$, 
$\cfrac { \log { x }  }{ \log { 3 }  } .\cfrac { \log { 2x }  }{ \log { x }  } .\cfrac { \log { y }  }{ \log { 2x }  } =2;\quad \quad \log { y } =2\log { 3 } =\log { 9 } ;\quad y=9$

$\log _{2x}216= x , (2x)^x = 216 , 2^x.x^x= 2^3 . 3^3$
An obvious solution to this eqquation is $x = 3$, so that (A) is the correct choice.

$\log _{ 3 }{ 9 } +\log _{ 5 }{ 25 } +\log _{ 2 }{ 8 } =\log _{ 3 }{ { \left( 3 \right)  }^{ 2 } } +\log _{ 5 }{ { \left( 5 \right)  }^{ 2 } } +\log _{ 2 }{ { \left( 2 \right)  }^{ 3 } } =2\log _{ 3 }{ { \left( 3 \right)  }^{  } } +2\log _{ 5 }{ { \left( 5 \right)  }^{  } } +3\log _{ 2 }{ { \left( 2 \right)  }^{  } } =7$

We have,

$(150)^x = 7$

We have,

$x=\log _a bc$

$x=\dfrac{\log bc}{\log a}$                     $\because \log _a m=\dfrac{\log m}{\log a}$

Similarly,

$y=\dfrac{\log ac}{\log b}$

$z=\dfrac{\log ab}{\log c}$

Since,

$\Rightarrow \dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}$

Therefore,

$\Rightarrow \dfrac{1}{\dfrac{\log bc}{\log a}+1}+\dfrac{1}{\dfrac{\log ac}{\log b}+1}+\dfrac{1}{\dfrac{\log ab}{\log c}+1}$

$\Rightarrow \dfrac{\log a}{{\log bc}+{\log a}}+\dfrac{\log b}{{\log ac}+{\log b}}+\dfrac{\log c}{{\log ab}+{\log c}}$

$\Rightarrow \dfrac{\log a}{{\log abc}}+\dfrac{\log b}{{\log abc}}+\dfrac{\log c}{{\log abc}}$

$\Rightarrow \dfrac{\log a+\log b+\log c}{{\log abc}}$

$\Rightarrow \dfrac{\log abc}{{\log abc}}$

$\Rightarrow 1$

Hence, this is the answer.

$ 3^{\log _{4}{x}}=27$
$\Rightarrow { 3 }^{ \log _{ 4 }{ x }  }={ 3 }^{ 3 }$
Equating powers of $3$
$\log _{ 4 }{ x } =3$
Taking exponential of the above equation, we get
${ 4 }^{ 3 }=x$
$\therefore x=64$

We have,
${ 3 }^{ \log _{ 4 }{ 5 }  }-{ 5 }^{ \log _{ 4 }{ 3 }  }$
${ =5 }^{ \log _{ 4 }{ 3 }  }-{ 5 }^{ \log _{ 4 }{ 3 }  },$ ($\because { x }^{ \log _{ a }{ y }  }={ y }^{ \log _{ a }{ x }  }$)
$=0$

$\log _{ k }{ x } \log _{ 5 }{ k } =\log _{ x }{ 5 } \ \Rightarrow \dfrac { \log { x } \log { k }  }{ \log { k } \log { 5 }  } \left[ \because \log _ba=\dfrac{\log a}{\log b}\right]=\dfrac { \log { 5 }  }{ \log { x }  } \ \Rightarrow { \left( \log { x }  \right)  }^{ 2 }={ \left( \log { 5 }  \right)  }^{ 2 }\ \Rightarrow \log { x } =\pm \log { 5 }=\log 5 , \log 5^{-1}, \left[n\log a=\log a^n \right] \ \therefore \quad x=5,\dfrac { 1 }{ 5 } $

Ans: B.C

Given equation is 
$4^{\log _{2}\log x}=\log x-\left ( \log x \right )^{2}+1$

Now, $\log _{2}\log x$ is meaningful if $\log x > 0$.
Since $4^{\log _{2}\log x}=2^{2\log _{2}\log x}=\left ( 2^{\log _{2}\log x} \right )^{2}=\left ( (\log x)^{\log _{2}2} \right )^{2}[\because a^{\log _bc}=c^{\log _ba}]$
             $=\left ( \log x \right )^{2}$    $ \left ( \because a^{\log _{a}x}=x, a> 0, a\neq 1 \right )$

So the given equation reduces to
$2\left ( \log x \right )^{2}-\log x-1=0$.
$\displaystyle \Rightarrow \log x=1, \log x=-\frac{1}{2}$.
But $\log x> 0$
Hence,    $\log x=1$, i.e.,$x=e$

Let $x=\left( \log _{ b }{ a }  \right) \left( \log _{ c }{ b }  \right) \left( \log _{ a }{ c }  \right) \ \Rightarrow x=\left( \dfrac { \log { a }  }{ \log { b }  }  \right) \left( \dfrac { \log { b }  }{ \log { c }  }  \right) \left( \dfrac { \log { c }  }{ \log { a }  }  \right),\left[\because \log _yx=\cfrac{\log x}{\log y}\right] \ \Rightarrow x=1$

Ans: C

$\displaystyle a^{\log _{a}10}= 10$
$\displaystyle \log _a10.\log _{10}a = 1\Rightarrow \cfrac{\log _{10}a}{\log _{10}a}=1$
Given expression is correct for all $'a'$ provided $'a'$ belongs to the domain of the given expression.
$\displaystyle \log _{a}10$ is defined for a+ive, and $\displaystyle a\neq 1.$
$\displaystyle \therefore a \in \left ( 0,1 \right )\cup \left ( 1,\infty  \right)$

Given 
$\displaystyle \log _{p}q+\log _{q}r+\log _{r}p=0$
$\Rightarrow \log _{p}q+\log _{q}r=-\log _{r}p$    ....(1)

Consider,$\left( \log _{ p } q \right) ^{ 3 }+\left( \log _{ q } r \right) ^{ 3 }+\left( \log _{ r } p \right) ^{ 3 }$

$=[(\log _{ p } q+\log _{ q } r)^{ 3 }-3\log _{ p } q\log _{ q } r(\log _{ p } q+\log _{ q } r)]+\left( \log _{ r } p \right) ^{ 3 }$

$=\left( -\log _{ r } p \right) ^{ 3 }-3\log _{ p } q\log _{ q } r\left( -\log _{ r } p \right) +\left( \log _{ r } p \right) ^{ 3 }$       (by (1))

$=3\log _{ p } q\log _{ q } r\log _{ r } p = 3 \quad \quad [\because \log _ab=\dfrac{\log b}{\log a}]$

$\displaystyle { \frac { 1 }{ 2 }  }^{ \log _{ 2 }{ 5 }  }={ 2 }^{ -\log _{ 2 }{ 5 }  }={ 2 }^{ \log _{ 2 }{ { 5 }^{ -1 } }  }={ 5 }^{ -1 }[\because a^{\log _ab}=b]=\frac { 1 }{ 5 } $

$\displaystyle 49^{A}= 49^{1-\log _{7}2}= \frac{49}{49^{\log _{7}2}}= \frac{49}{7^{\log _{7}4}}= \frac{49}{4}\quad [\because a^{\log _ab}=b]$
$\displaystyle 5^{B}= 5^{-\log _{5}4} \quad [\because m\log a=\log a^m]=5^{\log _5\dfrac{1}{4}}= \frac{1}{4}$
$\therefore 49^A+5^B = \cfrac{25}{2}$

$\displaystyle\frac{1}{1+\log _b\,a+\log _b\,c}+\displaystyle\frac{1}{1+\log _c\,a+\log _c\,b}+\displaystyle\frac{1}{1+\log _a\,b+\log _a\,c}$
$=\displaystyle \dfrac{1}{\displaystyle 1+\frac{\log a}{\log b}+\frac{\log c}{\log b}}+\dfrac{1}{\displaystyle 1+\frac{\log a}{\log c}+\frac{\log b}{\log c}}+\dfrac{1}{\displaystyle 1+\frac{\log b}{\log a}+\frac{\log c}{\log a}} [\because \log _yx=\dfrac{\log x}{\log y}]$
$=\displaystyle\frac{\log\,b}{\log\,b+\log\,a+\log\,c}+\displaystyle\frac{\log\,c}{\log\,c+\log\,a+\log\,b}+\displaystyle\frac{\log\,a}{\log\,c+\log\,a+\log\,b}\,=\,1$

Ans: D

We know $log _a  a = 1.         \therefore log _{25} 25 = 1$

Given, ${4.18}^{x} = 36.54$

$\log _8(64)$

$\log _4(64)$

$\log _2(64)$

$\log _{\sqrt3}81$

$\log _{\sqrt5}625$

We have,

${{\log } _{2a}}a=x,{{\log } _{3a}}2a=y,{{\log } _{4a}}3a=z$

$ {{\log } _{2a}}a=x $

$ \dfrac{\log a}{\log 2a}=x $

Similarly,

$ {{\log } _{3a}}2a=y $

$ \dfrac{\log 2a}{\log 3a}=y $

Similarly,

$ {{\log } _{4a}}3a=z $

$ \dfrac{\log 3a}{\log 4a}=z $

Therefore,,

$ =xyz-2yz $

$ =\dfrac{\log a}{\log 2a}\times \dfrac{\log 2a}{\log 3a}\times \dfrac{\log 3a}{\log 4a}-2\times \dfrac{\log 2a}{\log 3a}\times \dfrac{\log 3a}{\log 4a} $

$ =\dfrac{\log a}{\log 4a}-2\times \dfrac{\log 2a}{\log 4a} $

$ =\log a-\log 4-\log a-2\times \left( \log 2a-\log 4a \right) $

 $ =-\log 4-2\times \left( \log 2a-\log 2a-\log 2 \right) $

 $ =-\log 4+2\log 2 $

 $ =-\log 4+\log 4 $

 $ =0 $

So, the value is 0.

$\log _{\sqrt2}16$

$\log _{\sqrt2}256$

$\log _{\sqrt6}216$

$\log _{10}{(a^{2}-15a)}=2$

$ \log _{ 2 }{ x } +\log _{ 4 }{ x } +\log _{ 64 }{ x } = 5$

$\displaystyle y={ a }^{ \frac { 1 }{ 1-\log _{ a }{ x }  }  }\Rightarrow \log _{ a }{ y } =\frac { 1 }{ 1-\log _{ a }{ x }  } $, taking log both sides on base 'a'

${ \log _{ 2 }{ \log { x }  }  }$ is meaningful if $x>1$
Since ${ 4 }^{ \log _{ 2 }{ \log { x }  }  }={ \left( { 2 }^{ \log _{ 2 }{ \log { x }  }  } \right)  }^{ 2 }={ \left( \log { x }  \right)  }^{ 2 }$         .......$\left[\because  a^{ \log _{ a }{ x }  }=x,a>0,a\neq 1 \right] $     

So the given equation {${ 4 }^{ \log _{ 2 }{ \log { x }  }  }=\log { x } -{ \left( \log { x }  \right)  }^{ 2 }+1$} reduces to
$2{ \left( \log { x }  \right)  }^{ 2 }-\log { x } -1=0$
$\Rightarrow \log { x } =1,\log { x } =-\dfrac12$
But for $x>1$, $\log { x } >0$
 so $\log { x } =1$ i.e., $x=e$

Ans: C

If $\log _{ 4 }{ \left( x \right)  } =12$ then,

${ a }^{ \log _{ 3 }{ 7 }  }=27,\quad { b }^{ \log _{ 7 }{ 11 }  }=49,\quad { c }^{ \log _{ 11 }{ 25 }  }=\sqrt { 11 } $
$S=({ a }^{ \log _{ 3 }{ 7 } })^{\log _{ 3 }{ 7 }  }+{ \left( { b }^{ \log _{ 7 }{ 11 }  } \right)  }^{ \log _{ 7 }{ 11 }  }+{ \left( { c }^{ \log _{ 11 }{ 25 }  } \right)  }^{ \log _{ 11 }{ 25 }  }$
$={ \left( 27 \right)  }^{ \log _{ 3 }{ 7 }  }+{ \left( 49 \right)  }^{ \log _{ 7 }{ 11 }  }+{ \left( \sqrt { 11 }  \right)  }^{ \log _{ 11 }{ 25 }  }$
$={ 3 }^{ 3\log _{ 3 }{ 7 }  }+{ 7 }^{ 2\log _{ 7 }{ 11 }  }+{ 11 }^{ { 1 }/{ 2 }\log _{ 11 }{ 25 }  }$
$={ 7 }^{ 3 }+{ 11 }^{ 2 }+{ 25 }^{ { 1 }/{ 2 } }$
$=469$
Sum of digits $=19$
Hence, C is correct.