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# CAT Logarithm Questions [Difficult] with Solutions for Practice

Question 1:
The value of $x$ satisfying $\log _ { 10 } \left( 2 ^ { x } + x - 41 \right) = x \left( 1 - \log _ { 10 } 5 \right)$is:

[1] 35

[2] 40

[3] 41

[4] 43

Answer & Solution
Option # 3

We have,

$\quad \log _ { 10 } \left( 2 ^ { x } + x - 41 \right) = x \left( 1 - \log _ { 10 } 5 \right)$

$\Rightarrow \quad \log _ { 10 } \left( 2 ^ { x } + x - 41 \right) = x \log _ { 10 } 2 = \log _ { 10 } \left( 2 ^ { x } \right)$

$\Rightarrow \quad 2 ^ { x } + x - 41 = 2 ^ { x } \Rightarrow x = 41 .$

Question 2:
If the product of the roots of the equation, $x ^ { \left( \frac { 3 } { 4 } \right) \left( \log _ { 2 } x \right) ^ { 2 } + \log _ { 2 } x - \left( \frac { 5 } { 4 } \right) } = \sqrt { 2 }$ is $\frac { 1 } { \sqrt [ b ] { a } }$ (where a, b $\in N$) then the value of $( a + b )$

[1] 0

[2] 1

[3] 18

[4] 19

Answer & Solution
Option # 4

Take log on both the sides with base 2

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

$\log _ { 2 } x = y$

$3 y ^ { 3 } + 4 y ^ { 2 } - 5 y - 2 = 0$

$\Rightarrow \quad 3 y ^ { 2 } ( y - 1 ) + 7 y ( y - 1 ) + 2 ( y - 1 ) = 0$

$\Rightarrow \quad ( y - 1 ) \left( 3 y ^ { 2 } + 7 y + 2 \right) = 0$

$\Rightarrow \quad ( y - 1 ) ( 3 y + 1 ) ( y + 2 ) = 0$

$\Rightarrow \quad y = 1$ or $y = - 2$ or $y = \frac { - 1 } { 3 }$

Therefore, $\mathrm { x } = 2 ; \frac { 1 } { 4 } ; \frac { 1 } { 2 ^ { 1 / 3 } } \Rightarrow \mathrm { x } _ { 1 } \mathrm { x } _ { 2 } \mathrm { x } _ { 3 } = \frac { 1 } { \sqrt [ 3 ] { 16 } }$

$\Rightarrow \mathrm { a } + \mathrm { b } = 19$

Question 3:
For $0 < \mathrm { a } \neq 1 ,$ find the number of ordered pair $( \mathrm { x } , \mathrm { y } )$ satisfying the equation $\log _ { \mathrm { a } } | \mathrm { x } + \mathrm { y } | = \frac { 1 } { 2 }$ and $\log _ { \mathrm { a } } \mathrm { y } - \log _ { \mathrm { a } } | \mathrm { x } | = \log _ { \mathrm { a } ^ { 2 } } 4$

[1] 0

[2] 1

[3] 2

[4] 4

Answer & Solution
Option # 3

We have $\log _ { \mathrm { a } ^ { 2 } } | \mathrm { x } + \mathrm { y } | = \frac { 1 } { 2 } \Rightarrow \quad | \mathrm { x } + \mathrm { y } | = \mathrm { a } \quad \Rightarrow \quad \mathrm { x } + \mathrm { y } = \pm \mathrm { a } \quad \ldots (1)$

Also, $\log \left( \frac { \mathrm { y } } { | \mathrm { x } | } \right) = \log _ { \mathrm { a } ^ { 2 } } 4 \Rightarrow \quad \mathrm { y } = 2 | \mathrm { x } | \quad \ldots(2)$

If $\mathrm { x } > 0 ,$ then $\mathrm { x } = \frac { \mathrm { a } } { 3 } , \mathrm { y } = \frac { 2 \mathrm { a } } { 3 }$

If $\mathrm { x } < 0 ,$ then $\mathrm { y } = 2 \mathrm { a } , \mathrm { x } = - \mathrm { a }$

Therefore, possible ordered pairs $= \left( \frac { \mathrm { a } } { 3 } , \frac { 2 \mathrm { a } } { 3 } \right)$ and $( - \mathrm { a } , 2 \mathrm { a } )$

Question 4:
For $\mathrm { N } > 1 ,$ the product $\frac { 1 } { \log _ { 2 } \mathrm { N } } \cdot \frac { 1 } { \log _ { \mathrm { N } } 8 } \cdot \frac { 1 } { \log _ { 32 } \mathrm { N } } \cdot \frac { 1 } { \log _ { \mathrm { N } } 128 }$ simplifies to

[1] $\frac { 3 } { 7 }$

[2] $\frac { 3 } { 7 \ln 2 }$

[3] $\frac { 3 } { 5 \ln 2 }$

[4] $\frac { 5 } { 21 }$

Answer & Solution
Option # 4

$\frac { 1 } { \log _ { 2 } N } \cdot \frac { 1 } { \log _ { N } 8 } \cdot \frac { 1 } { \log _ { 32 } N } \cdot \frac { 1 } { \log _ { N } 128 }$

=$\frac { \ln 2 } { \ln N } \cdot \frac { \ln N } { 3 \ln 2 } \cdot \frac { 5 \ln 2 } { \ln N } \cdot \frac { \ln N } { 7 \ln 2 }$

= $\frac { 5 } { 21 }$

Question 5:
If p is the smallest value of $\mathrm { x }$ satisfying the equation $2 ^ { \mathrm { x } } + \frac { 15 } { 2 ^ { \mathrm { x } } } = 8$ then the value of $4 ^ { \mathrm { p } }$ is equal to

[1] 9

[2] 16

[3] 25

[4] 1

Answer & Solution
Option # 1

$2 ^ { 2 x } - 8 \cdot 2 ^ { x } + 15 = 0$

$\Rightarrow \left( 2 ^ { x } - 3 \right) \left( 2 ^ { x } - 5 \right) = 0$

$\Rightarrow 2 ^ { x } = 3$ or $2 ^ { x } = 5$

Hence smallest x is obtained by equating $2 ^ { x } = 3 \Rightarrow x = \log _ { 2 } 3$

So, $\quad p = \log _ { 2 } 3$

Hence, $4 ^ { p } = 2 ^ { 2 \log _ { 2 } 3 } = 2 ^ { \log _ { 2 } 9 } = 9$

Question 6:
The sum of two numbers a and b is $\sqrt { 18 }$ and their difference is $\sqrt { 14 } .$ The value of $\log _ { b }$ a is equal to

[1] -1

[2] 2

[3] 1

[4] $\frac{1}{2}$

Answer & Solution
Option # 1

$a + b = \sqrt { 18 }$

$a - b = \sqrt { 14 }$

squaring  & subtract, we get $4 \mathrm { ab } = 4 \Rightarrow \mathrm { ab } = 1$

Hence number are reciprocal of each other $\Rightarrow \log _ { \mathrm { b } } \mathrm { a } = - 1$

Question 7:
The value of the expression $\left( \log _ { 10 } 2 \right) ^ { 3 } + \log _ { 10 } 8 \cdot \log _ { 10 } 5 + \left( \log _ { 10 } 5 \right) ^ { 3 }$ is

[1] rational which is less than 1

[2] rational which is greater than 1

[3] equal to 1

[4] an irrational number

Answer & Solution
Option # 3

$\log _ { 10 } 2 = a$ and $\log _ { 10 } 5 = b$

$\Rightarrow \quad a + b = 1 ; a ^ { 3 } + 3 a b + b ^ { 3 } = ?$

Now $( a + b ) ^ { 3 } = 1 \Rightarrow a ^ { 3 } + b ^ { 3 } + 3 a b = 1$

Question 8:
If $x = \frac { \sqrt { 10 } + \sqrt { 2 } } { 2 }$ and $y = \frac { \sqrt { 10 } - \sqrt { 2 } } { 2 } ,$ then the value of $\log _ { 2 } \left( x ^ { 2 } + x y + y ^ { 2 } \right) ,$ is equal to

[1] 0

[2] 2

[3] 3

[4] 4

Answer & Solution
Option # 3

$\log _ { 2 } \left( ( \mathrm { x } + \mathrm { y } ) ^ { 2 } - \mathrm { xy } \right)$

but $\quad \mathrm { x } + \mathrm { y } = \sqrt { 10 } ; \mathrm { x } - \mathrm { y } = \sqrt { 2 } ; \quad \mathrm { xy } = \frac { 10 - 2 } { 4 } = 2$

${{\log }_{2}}(10-2)={{\log }_{2}}8=3$

Question 9:
The value of $6 + \log _ { \frac { 3 } { 2 } } \left( \frac { 1 } { 3 \sqrt { 2 } } \sqrt { 4 - \frac { 1 } { 3 \sqrt { 2 } } \sqrt { 4 - \frac { 1 } { 3 \sqrt { 2 } } \sqrt { 4 - \frac { 1 } { 3 \sqrt { 2 } } \cdots } } } \right)$ is

[1] 1

[2] 2

[3] -4

[4] 4

Answer & Solution
Option # 4

Let $\sqrt { 4 - \frac { 1 } { 3 \sqrt { 2 } } \sqrt { 4 - \frac { 1 } { 3 \sqrt { 2 } } } } \ldots \ldots = \mathrm { t }$

$\sqrt { 4 - \frac { 1 } { 3 \sqrt { 2 } } \mathrm { t } } = \mathrm { t }$

$4 - \frac { 1 } { 3 \sqrt { 2 } } \mathrm { t } = \mathrm { t } ^ { 2 }$

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

$\Rightarrow 3 \sqrt { 2 } t ^ { 2 } + t - 12 \sqrt { 2 } = 0$

$t = \frac { - 1 \pm \sqrt { 1 + 4 \times 3 \sqrt { 2 } \times 12 \sqrt { 2 } } } { 2 \times 3 \sqrt { 2 } } = \frac { - 1 \pm 17 } { 2 \times 3 \sqrt { 2 } }$

$\mathrm { t } = \frac { 16 } { 6 \sqrt { 2 } } , \frac { - 18 } { 6 \sqrt { 2 } }$

$\mathrm { t } = \frac { 8 } { 3 \sqrt { 2 } } , \frac { - 3 } { \sqrt { 2 } }$ and $\frac { - 3 } { \sqrt { 2 } }$ is rejected

So, $6 + \log _ { 32 } \left( \frac { 1 } { 3 \sqrt { 2 } } \times \frac { 8 } { 3 \sqrt { 2 } } \right)$

$= 6 + \log _ { 32 } \left( \frac { 4 } { 9 } \right)$

$= 6 + \log _ { 32 } \left( \left( \frac { 2 } { 3 } \right) ^ { 2 } \right) = 6 - 2 = 4$

Question 10:
Suppose that $x < 0 .$ Which of the following is equal to $\left| 2 x - \sqrt { ( x - 2 ) ^ { 2 } } \right|$

[1] x-2

[2] 3x-2

[3] 3x+2

[4] -3x+2

Answer & Solution
Option # 4

$y = | 2 x - | x - 2 | | = | 2 x - ( 2 - x ) | = | 3 x - 2 |$ as $x < 0$ hence $y = 2 - 3 x$

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