Download important CAT Exponents Problems with Solutions PDF based on previously asked questions in CAT exam. Practice Exponents Problems with Solutions for CAT exam.<\/p>\n
Download CAT Questions On Exponents<\/a><\/p>\n 3 Months Crash Course for CAT<\/a><\/p>\n Download CAT Quant Questions PDF<\/a><\/p>\n Take a free mock test for CAT<\/a><\/p>\n Question 1:\u00a0<\/b>X and Y are the digits at the unit’s place of the numbers (408X) and (789Y) where X \u2260 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)?<\/p>\n a)\u00a09<\/p>\n b)\u00a010<\/p>\n c)\u00a011<\/p>\n d)\u00a012<\/p>\n e)\u00a0None of the above<\/p>\n Question 2:\u00a0<\/b>For how many values of $Y>0$ is $Y = \\log_{Y}{10}$?<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a0More than twice<\/p>\n Question 3:\u00a0<\/b>If $\\log_{10}{2} = 0.3010$, find the number of digits in $5^{100}$<\/p>\n a)\u00a058<\/p>\n b)\u00a070<\/p>\n c)\u00a031<\/p>\n d)\u00a0None of These<\/p>\n Question 4:\u00a0<\/b>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<\/p>\n a)\u00a011<\/p>\n b)\u00a010<\/p>\n c)\u00a09<\/p>\n d)\u00a08<\/p>\n Question 5:\u00a0<\/b>$\\log_{7}{\\frac{1}{343}}$ is?<\/p>\n a)\u00a0-3<\/p>\n b)\u00a0-1\/3<\/p>\n c)\u00a03<\/p>\n d)\u00a0None of these<\/p>\n Take a free mock test for CAT<\/a><\/p>\n Download CAT Syllabus PDF<\/a><\/p>\n Question 6:\u00a0<\/b>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:<\/p>\n a)\u00a05<\/p>\n b)\u00a06<\/p>\n c)\u00a07<\/p>\n d)\u00a08<\/p>\n e)\u00a0None of the above<\/p>\n Question 7:\u00a0<\/b>$\\log{8}$ = 0.9030, what is $\\log{4}$?<\/p>\n a)\u00a00.4520<\/p>\n b)\u00a00.6020<\/p>\n c)\u00a00.5039<\/p>\n d)\u00a0None of these<\/p>\n Question 8:\u00a0<\/b>If $10^{67}- 87$ is written as an integer in base 10 notation, what is the sum of digits in that integer? \u00b7<\/p>\n a)\u00a0683<\/p>\n b)\u00a0489<\/p>\n c)\u00a0583<\/p>\n d)\u00a0589<\/p>\n Question 9:\u00a0<\/b>If $\\log_{10}{2}$ = 0.3010, how many digits does $2^{200}$ have?<\/p>\n a)\u00a060<\/p>\n b)\u00a061<\/p>\n c)\u00a062<\/p>\n d)\u00a0None of these<\/p>\n Question 10:\u00a0<\/b>What is the digit in the unit\u2019s place of $2^{51}$?<\/p>\n a)\u00a02<\/p>\n b)\u00a08<\/p>\n c)\u00a01<\/p>\n d)\u00a04<\/p>\n Take a free CAT online mock test<\/a><\/p>\n Question 11:\u00a0<\/b>Which of the following statements is false?<\/p>\n a)\u00a0For all X>1, $\\log_X{X}$=1<\/p>\n b)\u00a0For all X>1, $\\log_X{1}$=0<\/p>\n c)\u00a0For all X>Y>1, $\\log_Y{X}$>$\\log_X{Y}$>0<\/p>\n d)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>What are the last two digits of $7^{2008}$?<\/p>\n a)\u00a021<\/p>\n b)\u00a061<\/p>\n c)\u00a001<\/p>\n d)\u00a041<\/p>\n e)\u00a081<\/p>\n Question 13:\u00a0<\/b>If $\\log_{3}{A}$ = x and $\\log_{12}{A}$ = y, what is $\\log_{A}{6}$?<\/p>\n a)\u00a0(x+y)\/2xy<\/p>\n b)\u00a0(x+y)\/x<\/p>\n c)\u00a0(y+2x)\/xy<\/p>\n d)\u00a0None of these<\/p>\n Question 14:\u00a0<\/b>The rightmost non-zero digit of the number $30^{2720}$ is<\/p>\n a)\u00a01<\/p>\n b)\u00a03<\/p>\n c)\u00a07<\/p>\n d)\u00a09<\/p>\n Question 15:\u00a0<\/b>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?<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a05<\/p>\n d)\u00a06<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n All numbers from 1 to 9 repeat their last digits over a cycle of 4. 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n If $Y<1, \\log _{10}{Y}$ is negative and the equation has no solution. $\\log _{10}{Y}$<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\log{5^{100}} = 100 * ( 1 – \\log{2} ) = 69.87$. 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The sum of digits should be 2. The \u00a0possibilities are 1000000001,1000000010,10000000100,..these 10 cases . Also additional 1 case where 20000000000. Hence option A .<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\log_{7}{\\frac{1}{343}}$ = – $\\log_{7}{343}$ = -3<\/p>\n CAT Online Most Trusted Courses<\/a><\/p>\n 6)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $(xxx)_{b}=x^3$ 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\log{8}$ = 3 $\\log{2}$ = 1.5 $\\log{4}$. So, $\\log{4}$ = 0.9039\/1.5 = 0.6020<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $10^{67}- 87$ = $9999….99913$ (total 67 digits)<\/p>\n Sum of digits =$65*9 + 1 + 3$ = 589<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\log_{10}{2^{200}}$ = 60.20. Hence, $2^{200}$ has 60+1 = 61 digits<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The last digit of powers of 2 follow a pattern as given below.<\/p>\n The last digit of $2^1$ is 2 The last digit of $2^5$ is 2 Hence, the last digit of $2^{51}$ is 8<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\log_Y{X}$ > 1 (Since X > Y) All three statements are true.<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $7^4$ = 2401 = 2400+1 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n If $\\log_{3}{A}$ = x then $\\log_{A}{3}$ = 1\/x $\\frac{1}{x}$ + $\\frac{1}{y}$ = 2 $\\log_{A}{6}$ 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Rightmost non-zero digit of $30^{2720}$ is same as rightmost non-zero digit of $3^{272}$.<\/p>\n 272 is of the form 4k.<\/p>\n All $3^{4k}$ end in 1.<\/p>\n => Right most non-zero digit is 1.<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\log_{5}{(2^{x}-7)} + \\log_{5}{(2^{x}-4)} = 2\\log_{5}{(2^{x}-6)}$
\n63 can be written as 4k+3.
\n85 can be written as 4k+1.
\nSome number’s third power should yield the first digit as some number’s first power.
\n$2^3$ will yield $8$ as the last digit (2 and 8 is a possible solution).X+Y = 10
\n$3^3$ will yield $7$ as the last digit (3 and 7 is a possible solution).X + Y = 10
\n$4^3$ will yield $4$ as the last digit and hence, can be eliminated.
\n$5$ and $6$ yield $5$ and $6$ respectively as the last digit for any power and hence, can be eliminated.
\n$7^3$ will yield $3$ as the last digit\u00a0(7 and 3 is a possible solution). X+Y=10.
\n$8^3$ will yield $2$ as the last digit. 8+2 =10.
\nAs we can see, X+Y = 10 in all the cases. Therefore, option\u00a0B is the right answer.<\/p>\n
\nSo, $Y>1$ and $Y^{Y}$ = 10.
\nThe function $Y^{Y}$ is a monotonically increasing function and $2^{2} < 10$ and $3^{3}>10$.
\nSo, there exists exactly one Y between 2 and 3 such that $Y = \\log _{10}{Y}$<\/p>\n
\nSo, $5^{100}$ has 69+1=70 digits.<\/p>\n
\n=> $xb^2+xb+x = x^3$
\n=> $b^2+b+1=x^2$
\nOn 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.<\/p>\n
\nThe last digit of $2^2$ is 4
\nThe last digit of $2^3$ is 8
\nThe last digit of $2^4$ is 6<\/p>\n
\nThe last digit of $2^6$ is 4
\nThe last digit of $2^7$ is 8
\nThe last digit of $2^8$ is 6<\/p>\n
\n$\\log_X{Y}$ < 1 (Since Y< X)
\n$\\log_Y{X}$, $\\log_X{Y}$ > 0 (Since X and Y are > 1)<\/p>\n
\nSo, any multiple of $7^4$ will always end in 01
\nSince 2008 is a multiple of 4, $7^{2008}$ will also end in 01<\/p>\n
\nIf $\\log_{12}{A}$ = y then $\\log_{A}{12}$ = 1\/y
\n$\\log_{A}{3}$ + $\\log_{A}{12}$ = $\\log_{A}{36}$<\/p>\n
\n$\\log_{A}{6}$ = (x+y)\/2xy<\/p>\n
\nSo, $(2^{x}-7)*( 2^{x}-4)=(2^{x}-6)^{2}$.
\nAssuming $2^{x} = a$,
\n$(a-7)(a-4) = (a-6)^{2}$ or a=8. So, x=3<\/p>\n