CAT Questions On Exponents

CAT Questions On Exponents

Download important CAT Exponents Problems with Solutions PDF based on previously asked questions in CAT exam. Practice Exponents Problems with Solutions for CAT exam.

Download CAT Questions On Exponents

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Question 1: X and Y are the digits at the unit’s place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit’s place of the numbers $(408X{)}^{63}$ and $(789Y{)}^{85}$ are the same. What will be the possible value(s) of (X + Y)?

a) 9

b) 10

c) 11

d) 12

e) None of the above

Question 2: For how many values of $Y>0$ is $Y={\log }_{Y}10$?

a) 0

b) 1

c) 2

d) More than twice

Question 3: If ${\log }_{10}2=0.3010$, find the number of digits in $5^{100}$

a) 58

b) 70

c) 31

d) None of These

Question 4: Suppose n is an integer such that the sum of digits on n is 2, and ${10}^{10}<n<{10}^{11}$. The number of different values of n is

a) 11

b) 10

c) 9

d) 8

Question 5: ${\log }_7\frac{1}{343}$ is?

a) -3

b) -1/3

c) 3

d) None of these

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Question 6: Consider the expression $(xxx{)}_b=x^3$, where b is the base, and x is any digit of base b. Find the value of b:

a) 5

b) 6

c) 7

d) 8

e) None of the above

Question 7: $\log 8$ = 0.9030, what is $\log 4$?

a) 0.4520

b) 0.6020

c) 0.5039

d) None of these

Question 8: If ${10}^{67}-87$ is written as an integer in base 10 notation, what is the sum of digits in that integer? ·

a) 683

b) 489

c) 583

d) 589

Question 9: If ${\log }_{10}2$ = 0.3010, how many digits does $2^{200}$ have?

a) 60

b) 61

c) 62

d) None of these

Question 10: What is the digit in the unit’s place of $2^{51}$?

a) 2

b) 8

c) 1

d) 4

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Question 11: Which of the following statements is false?

a) For all X>1, ${\log }_{X}X$=1

b) For all X>1, ${\log }_{X}1$=0

c) For all X>Y>1, ${\log }_{Y}X$>${\log }_{X}Y$>0

d) None of these

Question 12: What are the last two digits of $7^{2008}$?

a) 21

b) 61

c) 01

d) 41

e) 81

Question 13: If ${\log }_{3}A$ = x and ${\log }_{12}A$ = y, what is ${\log }_{A}6$?

a) (x+y)/2xy

b) (x+y)/x

c) (y+2x)/xy

d) None of these

Question 14: The rightmost non-zero digit of the number ${30}^{2720}$ is

a) 1

b) 3

c) 7

d) 9

Question 15: If ${\log }_5(2^x-7)$, ${\log }_5(2^x-6)$ and ${\log }_5(2^x-4)$ are in arithmetic progression, what is the value of x?

a) 3

b) 4

c) 5

d) 6

Answers & Solutions:

1) Answer (B)

All numbers from 1 to 9 repeat their last digits over a cycle of 4.
63 can be written as 4k+3.
85 can be written as 4k+1.
Some number’s third power should yield the first digit as some number’s first power.
$2^3$ will yield $8$ as the last digit (2 and 8 is a possible solution).X+Y = 10
$3^3$ will yield $7$ as the last digit (3 and 7 is a possible solution).X + Y = 10
$4^3$ will yield $4$ as the last digit and hence, can be eliminated.
$5$ and $6$ yield $5$ and $6$ respectively as the last digit for any power and hence, can be eliminated.
$7^3$ will yield $3$ as the last digit (7 and 3 is a possible solution). X+Y=10.
$8^3$ will yield $2$ as the last digit. 8+2 =10.
As we can see, X+Y = 10 in all the cases. Therefore, option B is the right answer.

2) Answer (B)

If $Y<1,{\log }_{10}Y$ is negative and the equation has no solution.
So, $Y>1$ and $Y^Y$ = 10.
The function $Y^Y$ is a monotonically increasing function and $2^2<10$ and $3^3>10$.
So, there exists exactly one Y between 2 and 3 such that $Y={\log }_{10}Y$

${\log }_{10}Y$

3) Answer (B)

$\log 5^{100}=100\ast (1–\log 2)=69.87$.
So, $5^{100}$ has 69+1=70 digits.

4) Answer (A)

The sum of digits should be 2. The  possibilities are 1000000001,1000000010,10000000100,..these 10 cases . Also additional 1 case where 20000000000. Hence option A .

5) Answer (A)

${\log }_7\frac{1}{343}$ = – ${\log }_{7}343$ = -3

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6) Answer (E)

$(xxx{)}_b=x^3$
=> $x{b}^2+xb+x=x^3$
=> $b^2+b+1=x^2$
On substituting b=1,and b=2, we get $x^2$ as $3$, and $7$. Since $3$ and $7$ are not perfect squares, we can infer that no number satisfies the given condition. Therefore, option E is the right answer.

7) Answer (B)

$\log 8$ = 3 $\log 2$ = 1.5 $\log 4$. So, $\log 4$ = 0.9039/1.5 = 0.6020

8) Answer (D)

${10}^{67}-87$ = $9999\dots .99913$ (total 67 digits)

Sum of digits =$65\ast 9+1+3$ = 589

9) Answer (B)

${\log }_{10}{2}^{200}$ = 60.20. Hence, $2^{200}$ has 60+1 = 61 digits

10) Answer (B)

The last digit of powers of 2 follow a pattern as given below.

The last digit of $2^1$ is 2
The last digit of $2^2$ is 4
The last digit of $2^3$ is 8
The last digit of $2^4$ is 6

The last digit of $2^5$ is 2
The last digit of $2^6$ is 4
The last digit of $2^7$ is 8
The last digit of $2^8$ is 6

Hence, the last digit of $2^{51}$ is 8

11) Answer (D)

${\log }_{Y}X$ > 1 (Since X > Y)
${\log }_{X}Y$ < 1 (Since Y< X)
${\log }_{Y}X$, ${\log }_{X}Y$ > 0 (Since X and Y are > 1)

All three statements are true.

12) Answer (C)

$7^4$ = 2401 = 2400+1
So, any multiple of $7^4$ will always end in 01
Since 2008 is a multiple of 4, $7^{2008}$ will also end in 01

13) Answer (A)

If ${\log }_{3}A$ = x then ${\log }_{A}3$ = 1/x
If ${\log }_{12}A$ = y then ${\log }_{A}12$ = 1/y
${\log }_{A}3$ + ${\log }_{A}12$ = ${\log }_{A}36$

$\frac{1}{x}$ + $\frac{1}{y}$ = 2 ${\log }_{A}6$
${\log }_{A}6$ = (x+y)/2xy

14) Answer (A)

Rightmost non-zero digit of ${30}^{2720}$ is same as rightmost non-zero digit of $3^{272}$.

272 is of the form 4k.

All $3^{4k}$ end in 1.

=> Right most non-zero digit is 1.

15) Answer (A)

${\log }_5(2^x-7)+{\log }_5(2^x-4)=2{\log }_5(2^x-6)$
So, $(2^x-7)\ast (2^x-4)=(2^x-6{)}^2$.
Assuming $2^x=a$,
$(a-7)(a-4)=(a-6{)}^2$ or a=8. So, x=3

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