# CAT Questions On Exponents

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## 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.

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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

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

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.

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 $Y = \log _{10}{Y}$

$\log _{10}{Y}$

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

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 .

$\log_{7}{\frac{1}{343}}$ = – $\log_{7}{343}$ = -3

$(xxx)_{b}=x^3$
=> $xb^2+xb+x = x^3$
=> $b^2+b+1=x^2$
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.

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

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

Sum of digits =$65*9 + 1 + 3$ = 589

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

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

$\log_Y{X}$ > 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.

$7^4$ = 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

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

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.

$\log_{5}{(2^{x}-7)} + \log_{5}{(2^{x}-4)} = 2\log_{5}{(2^{x}-6)}$
So, $(2^{x}-7)*( 2^{x}-4)=(2^{x}-6)^{2}$.
Assuming $2^{x} = a$,
$(a-7)(a-4) = (a-6)^{2}$ or a=8. So, x=3