Tag: logarithms

Questions Related to logarithms

$\log \sqrt{6} {216}$ is equal to____

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $8$

  2. $4$

  3. $5$

  4. $6$

Show answer
Correct Option: D
Explanation:

$\log _{\sqrt6}216$

$=\cfrac1{\frac12}\log _6216$
$=2\log _66^3$
$=2\times3\log _66$
$=6$
Hence, D is the correct option.

The set of solutions for the equation $\log _{ 10 }{ \left( { a }^{ 2 }-15a \right)  } =2$ consists of:

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. Two integers

  2. One integer and one fraction

  3. two irrational numbers

  4. two non-real numbers

  5. no numbers, that is, the set is empty

Show answer
Correct Option: A
Explanation:

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

$\Rightarrow a^{2}-15a=10^{2}$
$\Rightarrow a^{2}-15a-100=0$
Discrimnant$= (15)^{2}-(4)(-100)=225+400= 625>0$
$(a-20)(a+5)=0$
$a=20,-5$
(Two integers)

If $\log _{2}x + \log _{4}x + \log _{64} x = 5$, then the value of $x$ will be

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $8$

  2. $16$

  3. $7$

  4. $2$

Show answer
Correct Option: A
Explanation:

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


We know that,
 
$ \log _{ b }{ a }=\dfrac { \log _{c} { a }  }{  \log _{c} {  b }   } $

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


 Assuming base $2$ for numerator and denominator, we get

$\dfrac { \log x }{ 1 } +\dfrac { \log x }{ 2 }  +\dfrac { \log x }{ 6 }  = 5$

$ 6\log x + 3\log x+\log x = 30$

$ 10 \log x = 30$

$ \log _{ 2 }{ { x }^{ 10 } }  =30 \ \ \ \ \dots (\log _{  { y}  }x^n=n \log _{ y }x)$

$ \\ { x }^{ 10 } = { 2 }^{ 30 }$

So, $x = { 2 }^{ 3 }=8$

If the eccentricity of the ellipse $\cfrac { { x }^{ 2 } }{ { \left( \log { a }  \right)  }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { \left( \log { b }  \right)  }^{ 2 } } =1\left( a>b>0,a\neq 1 \right) $ is $\cfrac { 1 }{ \sqrt { 2 }  } $ and $c$ be the eccentricity of the hyperbola $\cfrac { { x }^{ 2 } }{ { \left( \log _{ b }{ a }  \right)  }^{ 2 } } -{ y }^{ 2 }=1\quad $ then ${e}^{2}$ is greater than (where $\log{x}-\ln{x}$)

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $\dfrac{3}{2}$

  2. $\dfrac{1}{2}$

  3. $\dfrac{2}{3}$

  4. $\dfrac{5}{4}$

Show answer
Correct Option: A
Explanation:
Eqn of ellipse
$\dfrac{x^2}{(\log a)^2}+\dfrac{y^2}{(\log b)^2}=1\left(a>b>0, a\neq 1\right)$
if $a>b>0$ then $\log a> \log b$
For ellipse
$e=\dfrac{1}{\sqrt{2}}$
$e=\sqrt{1-\left(\dfrac{\log b}{\log a}\right)^2}=\dfrac{1}{\sqrt{2}}\quad \begin{cases} as\  & \dfrac { x^{ 2 } }{ a^{ 2 } } +\dfrac { y^{ 2 } }{ b^{ 2 } } =1 \\ e=\sqrt { 1-\dfrac { b^{ 2 } }{ a^{ 2 } }  }  & if\quad a>b \end{cases}$
$1-\left(\dfrac{\log b}{\log a}\right)^2=\dfrac{1}{2}$
$\dfrac{1}{2}=\left(\dfrac{\log b}{\log a}\right)^2\quad ----(1)$
Eqn of hyperbola
$\dfrac{x^2}{(\log _b a)^2}-y^2=1$
$e=\sqrt{1+\dfrac{b^2}{a^2}}\quad \left\{\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\right\}$
So
$e=\sqrt{1+\dfrac{1}{(\log _b a)^2}}=\sqrt{1+\left(\dfrac{\log b}{\log a}\right)}\quad \left\{ as\ \log _{ b }{ a=\dfrac { \log { a }  }{ \log { b }  }  }  \right. $
$=\sqrt{1+\dfrac{1}{2}}=\sqrt{\dfrac{3}{2}}$
$e^2=\dfrac{3}{2}$

If $\displaystyle y= a^{\left(\frac{1}{1-\log _{a}x}\right)}$ and $\displaystyle z= a^{\left(\frac{1}{1-\log _{a}y}\right)}$, then relation between $x$ and $z$ is

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $\displaystyle x= a^{\left(\frac{1}{1-\log _{a}z}\right)}$

  2. $\displaystyle x= a^{\left(\frac{1}{1+\log _{a}z}\right)}$

  3. $\displaystyle x= a\left(\frac{1}{1-\log _{a}z}\right)$

  4. $\displaystyle x= a\left(\frac{1}{1+\log _{a}z}\right)$

Show answer
Correct Option: A
Explanation:

$\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'

$\displaystyle z={ a }^{ \frac { 1 }{ 1-\log _{ a }{ y }  }  }\Rightarrow \log _{ a }{ z } =\frac { 1 }{ 1-\log _{ a }{ y }  } =\frac { 1 }{ 1-\frac { 1 }{ 1-\log _{ a }{ x }  }  } $

$\displaystyle \Rightarrow \log _{ a }{ z } =\frac { 1-\log _{ a }{ x }  }{ 1-\log _{ a }{ x } -1 } \Rightarrow -\log _{ a }{ x } \log _{ a }{ z } =1-\log _{ a }{ x } $

$\displaystyle \Rightarrow \log _{ a }{ x } =\frac { 1 }{ 1-\log _{ a }{ z }  } \Rightarrow x={ a }^{ \frac { 1 }{ 1-\log _{ a }{ z }  }  }$

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

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $x=1$

  2. $x=4$

  3. $x=e$

  4. $x={e}^{2}$

Show answer
Correct Option: C
Explanation:

${ \log _{ 2 }{ \log { x }  }  }$ 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\quad $, then $\log _{ 2 }{ \left( x/4 \right)  } $ is equal to

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $11$

  2. $48$

  3. $-12$

  4. $22$

Show answer
Correct Option: D
Explanation:

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


$x={ 4 }^{ 12 }\$ 

$\frac { x }{ 4 } ={ 4 }^{ 11 }$

Hence, 

$\log _{ 2 }{ \left( \frac { x }{ 4 }  \right)  } =\log _{ 2 }{ \left( { 4 }^{ 11 } \right)  } \$ 

                 $=\log _{ 2 }{ \left( \left( { 2 }^{ 2 } \right) ^{ 11 } \right)  } \$ 

                 $=\log _{ 2 }{ \left( { 2 }^{ 22 } \right)  }$ 

                 $ =22$

Hence option D is correct.

If $a, b, c$ are positive numbers such that $a^{\log _37}=27, b^{\log _711}=49, c^{\log _{11}25}=\sqrt{11}$, then the sum of digits of $S=a^{(\log _37)^2}+b^{(\log _711)^2}+c^{(\log _{11}25)^2}$ is

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. 15

  2. 17

  3. 19

  4. 21

Show answer
Correct Option: C
Explanation:

${ 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.