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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If `a gt 1, x gt 0` and `2^(log_(a)(2x))=5^(log_(a)(5x))`, then x is equal toA. `(1)/(10)`B. `(1)/(5)`C. `(1)/(2)`D. 1 |
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Answer» Correct Answer - A We having `2^(log_(a)(2x))=5^(log_(a)(5x))` Taking log on both sides, we get `log_(a)(2x).log 2=log_(a)(5x).log 5` `rArr ((log 2+log x))/(log a)log 2=((log 5+ log x))/(log a)log 5` `rArr (log 2)^(2)+log x log 2=(log 5)^(2)+(log x)log 5` `rArr log x(log 2-log 5)=(log 5)^(2)-(log 2)^(2)` `rArr -log x=log 5+log 2=log 10` `rArr x=(1)/(10)` |
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| 2. |
If `x_(1)` and `x_(2)` are solution of the equation `log_(5)(log_(64)|x|+(25)^(x)-(1)/(2))=2x`, thenA. `x_(1)=2x_(2)`B. `x_(1)+x_(2)=0`C. `x_(1)=3x_(2)`D. `x_(1)x_(2)=64` |
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Answer» Correct Answer - B `log_(5)(log_(64)|x|+(25)^(x)-(1)/(2))=2x` `rArr log_(64)|x|+(25)^(x)-(1)/(2)=(25)^(x)` `rArr log_(64)|x|=(1)/(2)` `rArr |x|=8` `rArr x=-8,8` |
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| 3. |
The value of x satisfying `5^logx-3^(logx-1)=3^(logx+1)-5^(logx - 1)` , where the base of logarithm is 10 is not : 67 divisible by |
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Answer» Correct Answer - 100 `5^(log x)-3^(log x-1)=3^(log x+1)-5^(log x-1)` `rArr 5^(log x)-3^(log x-1)=3^(log x+1)` `rArr 5^(log x)+5^(log x-1)=3^(log x+1)+3^(log x-1)` `rArr 5^(log x)+(5^(log x))/(5)=3.3^(log x)+(3^(log x))/(3)` `rArr (6.5^(log x))/(5)=(10.3^(log x))/(3)` `rArr ((3)/(5))^(log x)=((3)/(5))^(2)` `rArr log_(10)x=2` `rArr x =100` |
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| 4. |
`log_(3/4)log_8(x^2+7)+log_(1/2)log_(1/4)(x^2+7)^(-1)=-2`. |
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Answer» Correct Answer - `pm3` `log_(3//4)log_(8)(x^(2)+7)+log_(1//2)log_(1//4)(x^(2)+7)^(-1)=-2` `rArr log_(3//4)((1)/(3)log_(2)(x^(2)+7))-log_(2)(log_(2)(x^(2)+7))/(2)=-2` Let `log_(2)(x^(2)+7)=t` `rArr "log"_(3//4)(t)/(3)-"log"_(2)(t)/(2)+2=0` `rArr "log"_(3//4)(t)/(3)+1-("log"_(2)(t)/(2)-1)=0` `rArr "log"_(3//4)(t)/(4)="log"_(2)(t)/(4)` `rArr (t)/(4)=1` `rArr t = 4` `rArr log_(2)(x^(2)+7)=4` `rArr x^(2)+7=16` `rArr x^(2)=9` `rArr x^(2)= pm 3` |
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| 5. |
Solve:`log_axlog_a(xyz)=48`;`log_aylog_a(xyz)=12`;`log_azlog_a(xyz)=84` |
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Answer» Correct Answer - `x=a^(4),y=a,z=a^(7)` Adding given equations `log_(a)(xyz)[log_(a)x +log_(a)y+log_(a)z]=144` `rArr log_(a)(xyz)=(144)^(1//2)=12` `rArr xyz = a^(12)` From `log_(a)x log_(a)(xyz)=48` `rArr (log_(a)x)(12)=48` `rArr log_(a)x=4` `rArr =a^(4)` Similary y = a and `z=a^(7)` |
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| 6. |
Solve : `root(4)(|x-3|^(x+1))=root(3)(|x-3|^(x-2))`. |
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Answer» Correct Answer - x = 4, 2 ; x = 11 `root(4)(|x-3|^(x+1))=root(3)(|x-3|^(x-2))` Taking log on both the sides `(x+1)/(4)log|x-3|=(x-2)/(3)log|x-3|` `rArr log|x-3|[(x+1)/(4)-(x-2)/(3)]=0` `rArr log|x-3|=0` or `[((x+1)/(4))-((x-2)/(3))]=0` `rArr x=4, 2` or x = 11 |
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| 7. |
Which of the following is/are true ?A. number of digits in `8^(12)5^(35)` is 35B. number of digits in `8^(12)5^(35)` is 36C. number of zeroes after decimal before a significant figures starts in `((8)/(27))^(20)` is 10D. number of zeroes after decimal before a significant figure starts in `((8)/(27))^(20)` is 11 |
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Answer» Correct Answer - B::C Let `N_(1)=8^(12)5^(35)` `rArr log_(10)N_(1)=log_(10)(2^(36).5^(35))` `log_(10)(10^(35).2)` `=35+log_(10)2` `=35+0.3010` `= 35.3010` `therefore` Number of digits in `N_(1)` is 36. Now, let `N_(2)=((8)/(27))^(20)` `rArr log_(10)N_(2)=20 log_(10)((8)/(27))` `=60 "log"_(10)(2)/(3)` `=60(log_(10)2-log_(10)3)` `=60xx(0.3010-0.4771)` `=-60xx(0.3010-0.4771)` `-60xx0.1761` `=-10.56` `=bar(11)+0.44` `therefore` Number of zeroes after decimal before a significant figure starts in `N_(2)` is 10 |
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| 8. |
If `log_(3)x-(log_(3)x)^(2)le(3)/(2)log_((1//2sqrt(2)))4`, then x can belong toA. `(-oo,1//3)`B. `(9,oo)`C. (1,6)D. `(-oo,0)` |
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Answer» Correct Answer - A::B `log_(3)x-(log_(3)x)^(2)le(3)/(2)log_((1//2sqrt(2))4` `rArr log_(3)x-(log_(3)x)^(2)le -2` `rArr (log_(3)x)^(2)-log_(3)x-2 ge 0` `rArr (log_(3)x-2)(log_(3)x+1)ge 0` `rArr log_(3)x le -1` or `log_(3)x le 2` `rarr le 1//3` or `x ge 9` |
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| 9. |
The number of integers satisfying `log_((1)/(x))((2(x-2))/((x+1)(x-5)))ge 1` is |
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Answer» Correct Answer - A Case I: `1//x gt 1` or `0 lt x lt 1` `therefore log_((1)/(x))((2(x-2))/((x+1)(x-5)))ge 1` `rArr (2(x-2))/((x+1)(x-5))ge(1)/(x)` `rArr (2(x-2))/((x+1)(x-5))-(1)/(x)ge0` `rArr (2x(x-2)-(x+1)(x-5))/(x(x+1)(x-5))ge0` `rArr (x^(2)+5)/(x(x+1)(x-5))ge 0` `rArr x in (-1,0) uu (5, oo)` Hence, no solution in this case. Case II : `0 lt (1)/(x)lt 1` or `x gt 1` `therefore 0 lt x lt 5` Also `(2(x-2))/((x+1)(x-5))gt 0` `therefore x in (1,2)` |
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| 10. |
The value of `6^(log_10 40)*5^(log_10 36)` isA. 200B. 216C. 432D. none of these |
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Answer» Correct Answer - B We have `N = 6^(log_(10)40).5^(log_(10)36)` `therefore log_(10)N=log_(10)40 log_(10)6+log_(10)6+log_(10)36 log_(10)5` `=log_(10)6[log_(10)40+log_(10)25]` `= log_(10)6[log_(10)1000]` `= log_(10)(6)^(3)` `therefore N = 6^(3)=216` |
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| 11. |
The sum of all the values of a satisfying the equation `|[log_10 a,-1],[log_10(a-1),2]|=log_10 a+log_10 2` |
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Answer» Correct Answer - C `2log_(10)a+log_(10)(a-1)=log_(10)2a` `therefore a^(2)(a-1)=2a` `therefore a^(2)-a-2=0` `rArr (a-2)(a+1)=0` `rArr a=2,-1` (not possible) `therefore a = 2` |
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| 12. |
The number of real solution(s) of the equation `9^(log_(3)(log_(e )x))=log_(e )x-(log_(e )x)^(2)+1` is equal to |
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Answer» Correct Answer - B We have `9^(log_(3)(log_(e )x))=log_(e )x-(log_(e )x)^(2)+1` `therefore gt 1` The given equation `2(log_(e )x)^(2)-(log_(e )x)-1=0` `therefore log_(e ),(1)/(sqrt(e ))` (not possible) `rArr x = e` |
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| 13. |
The solution set of the equation `x^(log_x(1-x)^2)=9` isA. `{-2,4}`B. {4}C. `{0,-2,-4}`D. none of these |
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Answer» Correct Answer - B `x^(log_(x)(1-x)^(2))=9` `rArr (1-x^(2)=9` `rArr 1-x= pm 3` `rArr x=-2, 4` But x = -2 is not possible `rArr x = 4` |
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| 14. |
Let a and b be real numbers greater than 1 for which there exists a positive real number c, different from 1, such that `2(log_a c +log_b c)=9log_ab c`. Find the largest possible value of `log_a b`. |
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Answer» Correct Answer - 2 `a gt 1, b gt 1` `2(log_(a)c + log_(b)c)=9 log_(ab)c` `rArr 2[log c[(log b+log a)/(log a log b)]]=0 (log c)/(log a + log b)` `rArr 2(log a+ log b)^(2) = 9(log a)(log b)` `rArr 2(log a)^(2)+2(log b)^(2)+4(log a)(log b)= 9(log a)(log b)` `rArr 2log_(b)a + 2log_(a)b=5` `rArr t+(1)/(t)=(5)/(2)`, where `t=log_(a)b` `rArr 2r^(2)-5t+2=0` `rArr (2t-1)(t-2)=0` `rArr t = 1//2, t=2` `rArr log_(a)b=1//2` or `log_(a)b=2` |
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| 15. |
Solve : `log_(x^(2)16+log_(2x)64=3`. |
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Answer» Correct Answer - `x=2^(-1//3),4` `log_(x^(2))16+log_(2x)64=3` `rArr (4)/(log_(2)x^(2))+(6)/(log_(2)x)=3` Put `log_(2)x=t` `rArr (2)/(t)+(6)/(1+t)=3` `rArr 2+2t+6t=3t+3t^(2)` `rArr 3t^(2)-5t-2=0` `rArr (3t+1)(t-2)=0` `rArr t=-(1)/(3), t=2` `rArr log_(2)x=-(1)/(3)` or `log_(2)x=2` `rArr x=2^(-1//3)` or x = 4 |
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| 16. |
Solve : `log_(3)x . log_(4)x.log_(5)x=log_(3)x.log_(4)x+log_(4)x.log_(5)+log_(5)x.log_(3)x`. |
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Answer» Correct Answer - 60 `log_(3)x.log_(4)x.log_(5)x=log_(3)x.log_(4)x+log_(4)x.log_(5)x+log_(5)x.log_(3)x.` Let `log_(x)3=p, log_(x)4=q, log_(x)5=4` `rArr (1)/(pqr)=(1)/(pq)+(1)/(qr)=(1)/(pr)` `rArr p+q+r=1` `rArr log_(x)3+log_(x)4+log_(x)5=1` `rArr log_(x)60=1` `rArr x = 60` |
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| 17. |
Solve : `(3)/(2)log_(4)(x+2)^(2)+3=log_(4)(4-x)^(3)+log_(4)(6+x)^(3)`. |
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Answer» Correct Answer - `x=2,1- sqrt(23)` `(3)/(2)log_(4)(x+2)^(2)+3=log_(4)(4-x)^(3)+log_(4)(6+x)^(3)` `rArr 3=log_(4)((24-2x-x^(2))/(|x+2|))^(3)` `rArr ((24-2x-x^(2))/(|x+2|))^(3)=4^(3)` `rArr (24-2x-x^(2))/(|x+2|)=4` If `x+2gt0` `rArr x^(2)+6x-16=0` `rArr (x-2)(x+8)=0` `rArr x = 2` If `x+2 lt 0` `x^(2)-2x-32 =0` `rArr x=1 - sqrt(33)` |
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| 18. |
Solve `log_(6) 9-log_(9) 27 + log_(8)x = log_(64) x - log_(6) 4`..A. `1//2`B. `1//4`C. `1//8`D. `1//16` |
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Answer» Correct Answer - C `log_(6)9-log_(9)27)+log_(8)x=log_(64)x-log_(6)4` `rArr (log_(9)9+log_(6)4)-(3)/(2)log_(3)3=(log_(8)x)/(2)-log_(8)x` `rArr 2-(3)/(2)=-(1)/(2)log_(8)x` `rArr x=(1)/(8)` |
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| 19. |
If `log_(2)(log_(2)(log_(2)x))=2`, then the number of digits in x, is `(log_(10)2=0.3010)`A. 7B. 6C. 5D. 4 |
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Answer» Correct Answer - C `log_(2)(log_(2)(log_(2)x))=2` `rArr log_(2)(log_(2)x)=4` `rArr log_(2)x=16` `rArr x=2^(16)` `therefore log_(10)x=16log_(10)2=16xx0.3010=4.8160` `therefore` Number of digits = 5 |
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| 20. |
The smallest integral x satisfying the inequality `(1-log_(4)x)/(1+log_(2)x)le (1)/(2)x is.A. `sqrt(2)`B. 2C. 3D. 4 |
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Answer» Correct Answer - B Let `log_(2)x=t` `therefore (1-(t//2))/(1+t)le(1)/(2)` `rArr (2-t)/(1+t)le 1` `rArr (2-t)/(1+t)-1le 0` `rArr (2t-1)/(t+1)ge 0` `rArr t lt -1` or `t ge (1)/(2)` `rArr log_(2)x gt-1` or `log_(2)x ge (1)/(2)` `rArr 0 lt x lt (1)/(2)` or `x ge sqrt(2)` `rArr` smallest integer is 2 |
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