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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.
| 51. |
Solve : `6(log_x2-(log_4x)+7=0.` |
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Answer» `6(log_x2 - log_4x)+7 = 0` `=>6(log_2 2/(log_2 x) - log_2x/log_2 4) +7 = 0` `=>6(1/(log_2 x) - log_2x/2) +7 = 0` Let `log_2x =t`, then our equation becomes, `=>6(1/t - t/2)+7 =0` `=>12-6t^2+14t = 0` `=>3t^2- 7t -6 =0` `=>3t^2- 9t + 2t -6 =0` `=>(t-3)(3t+2) = 0` `=>t = 3 or t = -2/3` `=>log_2 x = 3 or log_2 x = -2/3` `=>x = 2^3 or x = 2^(-2/3).` |
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| 52. |
Solve for x: `4^(x) - 3^(x-1/2)=3^(x+1/2)-2^(2x-1)`. |
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Answer» `4^(x)-3^(x-1/2)=3^(x+1/2) - 2 ^(2x-1)` ` or 4^(x) - 3^(x)/sqrt3 = 3^(x) sqrt3 - 4^(x)/2` ` or 3/2 xx 4^(x) = 3^(x) (sqrt3+ 1/sqrt3)` `or 3/2 xx 4^(x) = 3^(x) xx 4/sqrt3 ` `or 4^(x-1)/4^(1//2) = 3^(x-1)/sqrt3` ` or 4^(x-3//2)= 3^(x-3//2)` ` or (4/3)^(x-3//2) = 1` `or x-3/2 = 0` ` or x = 3//2` |
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| 53. |
Solve `e^(sin x)-e^(-sin x) - 4 = 0`. |
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Answer» Given that ` e^(sin x) - e^(-sin x) - 4 = 0`. Let ` e^(sin x) = y`. Then, given equation becomes ` y-1/y - 4 = 0` ` or y^(2) - 4y-1 = 0` `:. Y = e^(sin x) = 2 + sqrt5, 2- sqrt5` Since ` e^(sin x) gt 0, e^(sin x) ne 2 - sqrt5` Also, maximum value of ` e^(sin x )` is e when sin x = `. So, ` e^(sin x) ne 2+ sqrt5` Therefore, given equation has no solution. |
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| 54. |
Solve `|x-3|^(3x^(2)-10x+3)= 1`. |
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Answer» Given that `|x-3|^(3x^(2) -10x+3) = 1`, we have following cases If ` 3x^(2) - 10x + 3 = 0," where "x ne 3` ` :. X = 1//3` If ` |x-3| = 1, :. X = 4, 2` |
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| 55. |
Find the smallest integral value of x satisfying `(x-2)^(x^(2)-6x+8) gt 1`. |
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Answer» Clearly , ` x gt 2` …(i) `(x-2)^(x^(2)-6x+8) gt 1` ` rArr (x-2)^(x^(2)-6x+8) gt (x-2)^(0)` When ` x - 2 gt 1` ` or x gt 3`…(ii) We have ` x^(2) - 6x+8 gt 0 ` ` rArr (x-2)(x-4) gt 0 ` ` rArr x lt 2 or x gt 4 ` ....(iii) From (ii) and (iii) ,` x gt 4` When ` x - 2 lt 1` ` or x lt 3 ` ....(iv) We have ` x^(2) - 6x + 8 lt 0 ` ` :. (x-2)(x-4) lt 0` ` :. 2 lt x lt 4` ...(v) From (i), (iv) and (v), we have ` 2 lt x lt 3` Thus, `x in (2, 3) cup (4, infty)`. |
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| 56. |
Solve:`|x-3|^(3x^2-10 x+3)=1` |
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Answer» `|x-3|^(3x^2-10x+3)=1` taking log both side `(3x^2-9x-x+3)log(|x-3|)` `(3x(x-3)-1(x-3)log(|x-3|_=0` `(3x-1)(x-3)(log|x-3|)=0` `x=1/3,2,4`. |
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| 57. |
Find the smallest integral value of `x`satisfying `(x-2)^(x^2-6x+8))>1` |
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Answer» `(x-2)^(x^2-6x+8)` 1)`x-2<1` `x^2-6x+8>0` `x^2-4x-2x+8>0` `(x-4)(x-2)>0` `x in (-oo.2)(4,oo)` `x in(4,oo)` 2)`x-2<1` `x<3` `x-6x+8<0` `(x-4)(x-2)<0` `x in(2,4)` `x in(2,3)` `x in(2,-3)uu(4,p)` `5`. |
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| 58. |
The solution of the equation `(log)_7(log)_5(sqrt(x+5)+sqrt(x)=0`is... |
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Answer» `log_7 (log_5 (sqrt(x+5)+sqrtx))=0` `log_5(sqrt(x+5)+sqrtx))=7^0=1` `sqrt(x+5)+sqrtx=5^1=5` Squaring both sides `x+5+x+2sqrtxsqrt(x+5)=25` `2sqrtxsqrt(x+5)=25-5-2x` `2sqrtxsqrt(x+5)=20-2x=2(10-x)` `sqrtxsqrt(x+5)=10-x` Squaring both sides `x(x+5)=(10-x)^2` `x^2-5x=x^2-20x+100` `25x=100` `x=4` |
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| 59. |
Find the value of ` sqrt((log_(0.5)4)^(2))`. |
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Answer» Correct Answer - 2 `sqrt((log_(0.5)4)^(2)) = |log_(0.5) 4|` ` = |log_(0.5)(0.5)^(-2)|` ` |-2|` = 2 |
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| 60. |
If ` log_(sqrt8) b = 3 1/3`, then find the value of b. |
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Answer» Correct Answer - 32 ` log_(sqrt8) b = 3 1/3` ` rArr b = (sqrt8)^(10/3) = 2^(5) = 32` |
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| 61. |
Find the value of ` log_(5) log_(2)log_(3) log_(2) 512`. |
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Answer» ` log_(5) log_(2) log_(3) log_(2) (2^(9))` ` = log_(5) log_(2) log_(3) 9` ` = log_(5) log_(2) log_(3) 3^(2) ` ` = log_(5) log_(2) 2` ` = log_(5) 1` = 0 |
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| 62. |
If the equation `2^x+4^y=2^y`is solved for `y`in terms of `x`where `x |
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Answer» Correct Answer - B `2^(2y) - 2^(y) + 2^(x) (1-2)^(x)) = 0` Putting ` 2^(y) = t`, we get ` t^(2) - t+2^(x) (1-2^(x)) = 0 ," where "t_(1) = 2^(y_(1)) and t_(2) = 2^(y_(2))` ` t_(1)t_(2)=2^(x)(1-2^(x))` ` 2^(y_(1)+y_(2))=2^(x) (1-2^(x))` ` y_(1)+y_(2) = x + log _(2) (1-2^(x))` |
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| 63. |
The number of solution of `x^(log_(x)(x+3)^(2)) = 16`is |
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Answer» Correct Answer - A `x^(log_(x)(x+3)^(2))=16," where " x gt x ne 1` ` rArr (x+3)^(2) = 16 ` ` or x =1, -7` Therefore, both values are not possible. |
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| 64. |
If x = 9 is one of the solutions of ` log_(e)(x^(2)+15a^(2))-log_(e)(a-2)=log_(e)((8ax)/(a-2))`,thenA. ` a= 3/5`B. a = 3C. x = 15D. x = 2 |
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Answer» Correct Answer - B `log_(e)(x^(2)+15a^(2))-log_(e)(a-2)=log_(e)((8ax)/(a-2))` Here` a gt 2, (8ax)/(a-2) gt 0` `:. x gt 0` Now` (x^(2)+15a^(2))/(a-2) = (8ax)/(a-2)` ` or x^(2) = 8ax+15a^(2) = 0` ` rArr x = 3a, 5a` Now, ` x = 9` ` rArr a = 3, 9/5` But ` a gt 2` ` :. a = 3` For ` a = 3, x = 9, 15` |
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| 65. |
The product of roots of the equation `(log_(8)(8//x^(2)))/((log_(8)x)^(2)) = 3` isA. 1B. `1//2`C. `1//3`D. `1//4` |
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Answer» Correct Answer - D Let ` log_(8) x = y`, then the given equation reduces to ` (1-2y)//y^(2) = 3`. ` rArr 3y^(2)+2y - 1 = 0` `(3y-1)(y+1) = 0` ` rArr log_(8) x = y = 1//3, -1` ` rArr x = 2, 1//8` |
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| 66. |
Let `a >1`be a real number. Then the number of roots equation `a^(2(log)_2x)=15+4x^((log)_2a)`is2 (b) infinite (c)0 (d) 1 |
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Answer» `a^(2log_2^x)=15+4x^(log_2^a)` `a>1` `(a^(log_2^x))^2=15+4x^(log_2^a)` `(x^(log_2^a))^2=15+4x^(log_2^a)` `x>0` Let `x^(log_2^a)=t` `log_2^a` is positive `t^2=15+4t` `t^2-4t-15=0` `t=(4pmsqrt(16+60))/2=(4pmsqrt(76))/2` `t=2+sqrt19` `t=2-sqrt19` this is not possible Number of solution=1. |
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| 67. |
Let `agt1` be a real number . If S is the set of real number `x` that are solutions to the equation `a^(2log_2x)=5+4x^(log_2a)`, thenA. 2B. infiniteC. 0D. 1 |
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Answer» Correct Answer - D Given equation can be written as `(a^(log_(2)x))^(2)=5+4 a^(log_(2) x)` Let `a^(log_(2)x)=t`.Then the given equation becomes ` r^(2) - 4t-5 = 0` `rArr (t-5)(t+1) = 0` ` rArr t = 5 or t =- 1` (rejected) ` :. A^(log_(2)x) = 5` ` rArr x^(log_(2)a) = 5` ` rArr x = 5^(log_(a)2)` |
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| 68. |
Solve : `(log)_2(x-1)/(x-2)>0` |
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Answer» `log_2((x-1)/(x-2)) gt 0` `=>(x-1)/(x-2)gt 2^0` `=>(x-1)/(x-2)gt 1` `=>(x-1)/(x-2) - 1gt 0` `=>((x-1)-(x-2))/(x-2) gt 0` `=>1/(x-2) gt 0` It will be possible, when `x gt 2`. So, `x in (2,oo)` is the solution for this equation. |
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| 69. |
Find the value of the following: (i) ` log_(10) 2 + log_(10) 5` (ii) ` log_(3) (sqrt(11)-sqrt2) + log_(3) (sqrt11+sqrt2)` (iii) ` log_(7) 35 - log_(7) 5` |
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Answer» (i) `log_(10)2+ log_(10) 5 = log_(10)(2 xx 5)= log_(10) 10 = 1` `(ii) log_(3)(sqrt11-sqrt2)+log_(3)(sqrt11+sqrt2)` ` " "= log_(3) (sqrt11-sqrt2) xx (sqrt11+sqrt2)` `" "= log_(3) (11-2)=log_(3)9=2 (as 3^(2) = 9)` `(iii) log_(7)35-log_(7) 5 = log _(7). 35/5 = log_(7) 7 = 1` |
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| 70. |
Solve `(log)_2(3x-2)=(log)_(1/2)x` |
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Answer» `log_(2)(3x-2)=log_(1//2)x=-log_(2)x=log_(2)x^(-1)` `or3x-2=x^(-1)or3x^(2)-2x=1` `rArrx=1orx=-1//3`. But `log_(2)(3x-2)` and `log_(1//2)x` are meaningful if `xgt2//3`. Hence, x=1 |
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| 71. |
` log_(x-1)x*log_(x-2)(x-1)*...*log_(x-12)(x-11) = 2`,x is equal toA. 9B. 16C. 25D. none of these |
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Answer» Correct Answer - B `log_((x-12)) x = 2` ` x = (x-12)^(2)` ` x ^(2) - 25_(x) + 144 = 0` ` x = 9, 16` but ` x ne 9, x = 16` |
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| 72. |
The value of `((log)_2 24)/((log)_(96)2)-((log)_2 192)/((log)_(12)2)`is3 (b) 0(c) 2 (d) 1A. 3B. 0C. 2D. 1 |
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Answer» Correct Answer - A Let ` log_(2) 12 =a`, then ` 1/(log_(96)2) = log_(2) 96 = log_(2) 2^(3) xx 12 = 3 + a` ` log_(2) 24=1+a` ` rArr log_(2) 192 = log_(2) (16xx12) = 4 + a` ` and 1 /(log_(12)2) = log_(2) 12 = a` . Therefore, the given expression is: ` (1+a)(3+a)-(4+a) a = 3` |
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| 73. |
The value of `((log)_2 24)/((log)_(96)2)-((log)_2 192)/((log)_(12)2)`is3 (b) 0(c) 2 (d) 1 |
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Answer» `[log_2^24*log_2^96]-[log_2^192]*[log_2^12]` `[log_2^(2^2*3)]*[log_2^(2^5^3)]*[log_2^(2^6*3)]*[log_2^(2^2*3)]` `[log_2^(2^3)+log_2^3][log_2^(2^5)+log_2^3)]-[log_2^(2^6)+log_2^3]*[log_2(2^2)+log_2^3]` `[3+log_2^3][5+log_2^3]-[6+log_2^3][2+log_2^3]` `15+8log_2^3+(log_2^3)^2-12-8log_2^3-(log_2^3)^2` `=15-12` `=3` Option A is correct. |
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| 74. |
Find the value of 7 ` log(16/15) + 5 log (25/24) + 3 log (81/80)`. |
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Answer» `7 log(16/15)+5 log(25/24)+3 log (81/80)` ` =log [(16/15)^(7)(25/24)^(5)(81/80)^(3)]` ` = log.2^(28)/(3^(7)5^(7))5^(10)/(2^(15)3^(5))3^(12)/(2^(12)5^(3))` ` = log 2` |
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| 75. |
Find the value of`(1/(49))^(1+(log)_7 2)+5^(-1(log)_((1/5))(7))` |
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Answer» `(1/49)^(1+log_7^2)` `1/7^2(1+log_7^2)` `7^2(1+log_7^2)` `7^2*7^(2log_7^2)` `7^2*7^(log_7^4)` `7^2*4` `1/(7^2*4)` `5^(-log_(1/5)^7` `7` Answer=`7+1/(4*7^2)`. |
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| 76. |
Prove that ` log_(10) 2" lies between " 1/4 and 1/3`. |
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Answer» We have to prove that`1/4 lt log_(10)2 lt 1/3`. Consider ` log_(10) 2 gt 1/4` ` rArr 2 gt 10^(1/4)` `rArr 2^(4) gt 10`, which is true Now consider ` log_(10) 2 lt 1/3` ` rArr 2 lt 10^(1/3)` ` rArr 2^(3) lt 10 `, which is true. Thus, ` 1/4 lt log_(10) 2 lt 1/3` |
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| 77. |
Consider the system of equations ` log_(3)(log_(2)x)+log_(1//3)(log_(1//2)y) =1 and xy^(2) = 9`. The value of 1/y lies in the intervalA. `(5, 7)`B. `(7, 10)`C. `(11, 15)`D. `(25, 30)` |
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Answer» Correct Answer - B `log_(3)(log_(2)x)+log_(1//3)(log_(1//2)y) = 1` ` rArr log_(3) (log_(2)x)-log_(3)(-log_(2)y) = 1` ` rArr log_(3)(-(log_(2)x)/(log_(2)y)) = 1` ` rArr -(log_(2)x)/(log_(2)y) = 3` ` rArr xy^(3) = 1` Also, ` xy^(2) = 9` ` rArr y = 1/9` ` :. x = 729` |
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| 78. |
Consider the system of equations ` log_(3)(log_(2)x)+log_(1//3)(log_(1//2)y) =1 and xy^(2) = 9`. The value of x in the intervalA. `(200, 300)`B. ` (400, 500)`C. `(700, 800)`D. none of these |
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Answer» Correct Answer - C `log_(3)(log_(2)x)+log_(1//3)(log_(1//2)y) = 1` ` rArr log_(3) (log_(2)x)-log_(3)(-log_(2)y) = 1` ` rArr log_(3)(-(log_(2)x)/(log_(2)y)) = 1` ` rArr -(log_(2)x)/(log_(2)y) = 3` ` rArr xy^(3) = 1` Also, ` xy^(2) = 9` ` rArr y = 1/9` ` :. x = 729` |
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| 79. |
If `log_(y) x + log_(x) y = 2, x^(2)+y = 12` , then the value of xy isA. 9B. 12C. 15D. 21 |
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Answer» Correct Answer - A Let ` t=log_(y) x(x, y gt 0, and ne 1)," then " t+1/t = 2 or (t-1)^(2) = 0` ` :. T = log_(y) x = 1 , i.e., x = y `.We get ` x^(2) + x - 12 = 0 x = -4, 3`. x = 3 only (-4 rejected) ` :. Y = 3`, ` :. X = 9` |
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| 80. |
If `xy^(2) = 4 and log_(3) (log_(2) x) + log_(1//3) (log_(1//2) y)=1` , then x equalsA. 4B. 8C. 16D. 64 |
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Answer» Correct Answer - D `log_(3)(log_(2)x)+log_(1//3)(log_(1//2)y)= 1` `or log_(3)(log_(2)x)-log_(3)(log_(1//2)y) = 1` ` or log_(3)(log_(2)(4//y^(2)))-log_(3)(log_(1//2)y) = 1` ` or log_(2)(4//y^(2))=3(log_(1//2)y)` ` or log_(2)(4//y^(2))=- 3 (log_(2)y)` `or log_(2)(4//y^(2))+(log_(2)y^(3))=0` ` or 4y = 1` ` or y = 1//4` ` rArr x = 64` |
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| 81. |
The least integer greater than ` log_(2) 15* log_(1//6 2* log_(3) 1//6` is _______. |
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Answer» Correct Answer - 3 ` log_(2) 15 * log_(1//6) 2* log_(3) 1//6` ` = (log 15)/(log 2) *(log 2)/(-log 6) * (-log 6)/(log 3) ` ` = log_(3) 15` |
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| 82. |
If `log_(4)A=log_(6)B=log_(9)(A+B)," then "[4(B//A)]("where "[*]` represents the greatest integer function ) equals _______. |
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Answer» Correct Answer - 6 Let ` log_(4)0A=log_(6)B=log_(9)(A+B) = x` `rArr A = 4^(x), B=6^(x) and A+B=9^(x)` ` A+B = 9^(x) rArr 4^(x) + 6^(x) = 9^(x)` ` rArr 2^(2x)+2^(x)*3^(x)=3^(2x)` `rArr (3/2)^(2x)-(3/2)^(x) -1=0` ` rArr(3/2)^(x) = (1+sqrt5)/2` ` rArr B/A = (3/2)^(x) = (1+sqrt5)/2` ` rArr 4B/A = 4((1+sqrt5)/2)` ` rArr [4B/A] = 6` |
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| 83. |
The value of `(log_(10)2)^(3)+log_(10)8 * log_(10) 5 + (log_(10)5)^(3)` is _______. |
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Answer» Correct Answer - 1 Let ` log_(10)2 = p and log_(10) 5 = q` Hence, p+q = 1 ` x = p^(3) + 3pq + q^(3)` ` = (p+q)^(3) - 3 pq(p+q)+3 pq` ` = 1-3pq+3pq` = 1 |
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| 84. |
If ` log_(a)b=2, log_(b)c=2, and log_(3) c= 3 + log_(3)` a,then the value of c/(ab)is ________. |
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Answer» Correct Answer - 3 ` log_(3) c = 3 + log _(3) a ` ` or log_(3) c/a = 3 or c = 27 a ` …(i) ` log_(a) b = 2, log_(b) c = 2` ` rArr log_(a) b* log _(b) c = 4` ` rArr log_(a) c = 4` ` rArr c = 4^(4)`...(ii) From Eqs.(i) and (ii), we get a = 3, c = 81. From ` log_(a) b = 2," we have " b = a^(2) = 9`. Hence, c/(ab) = 3. |
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| 85. |
If `log_a(ab)=x` then `log_b(ab)` is equals to |
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Answer» `log_(a) (ab) = x or log_(a) a+ log_(a) b = x or log_(a) b= x - 1` Now, ` log_(b)(ab) = log_(b) a+ log_(b) b` ` = 1/(log_(a) b) + 1 = 1/(x-1) + 1 x/(x-1)` |
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| 86. |
Prove that ` log_(7) log_(7)sqrt(7sqrt((7sqrt7))) = 1-3 log_(7) 2`. |
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Answer» `log_(7) log_(7) sqrt(7sqrt((7sqrt7))) = log_(7) log_(7) 7^(1/2+1/4+1/8)` ` = log_(7) (1/2+1/4+1/8)` ` = log_(7) (7/8)` ` = 1- log_(7) 8` ` = 1-3 log_(7) 2` |
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| 87. |
Solve for x:` 11^(4x-5)*3^(2x)=5^(3-x) * 7^(-x)`. |
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Answer» `11^(4x-5)*3^(2x)= 5^(3-x) * 7^(-x)*` ` or (4x-5)log 11+2xlog3 = (3-x)log 5-x log 7` ` or x= log(11^(5) xx 5^(3))/(log(11^(4) xx 315))` |
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| 88. |
If ` log_(10) x = y ," then find "log_(1000)x^(2)" in terms of " y`. |
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Answer» Correct Answer - `2/3 y` ` log_(1000)(x^(2)) = 2/3 log_(10) x =- 2/3 y` |
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| 89. |
Which is greater : `x=log_3 5or y= log_17 (25)` |
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Answer» `1/y = log_(25) 17 = 1/2 log_(5) 17 and 1/x = log_(5) 3 = 1/2log_(5) 9` `:. 1/y gt 1/x or x gt y` |
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| 90. |
If ` log_(a) 3 = 2 and log_(b) 8 = 3," then prove that "log_(a) b= log_(3) 4`. |
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Answer» If ` log_(a) 3 = 2` ` rArr 3 = a^(2)` ` rArr a = sqrt3` If ` log_(b) 8 = 3` ` rArr 8 = b^(3)` ` rArr b = 2` So, ` log_(a) b = log_(sqrt3) 2 = x(let)` ` rArr 2 = (sqrt3)^(x)` ` rArr 4 = 3^(x)` ` rArr x = log_(3) 4` |
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| 91. |
If characteristic of three numbers a, b and c and 5, -3 and 2, respectively, then find the maximum number of digits in N = abc. |
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Answer» Correct Answer - 7 According to the given information, ` log a = 5+p` ` log b =- 3+ q` ` log c = 2 + r` where p,q and r are mantissas. `:. P,q,r in [0, 1)` Adding the above equations, we get ` log(abc) = 4 + (p+q+r)` ` (p+q+r) in [0, 3)` ` :. log (abc) in [4, 7)` `rArr log N in [4, 7)` ` :. ` Maximum possible characteristic of log N = 6 `:. ` Maximum number of digits in `N = 6 +1 = 7` |
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| 92. |
Find the value of `log" " tan 1^(@) log" " tan 2^(@) … log" " tan 89^(@)`. |
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Answer» `log tan 1^(@) log tan 2^(@)* * * log tan 89^(@)` `" "=log tan 1^(@) log tan 2^(@)* * * log tan 45^(@) * * * log tan 89^(@)` `" "= log tan 1^(@) log tan 2^(@) * * * (log 1) * * * log tan 89^(@)` ` " "=0 (as log 1 = 0)` |
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| 93. |
There are 3 number a, b and c such that ` log_(10) a = 5.71, log_(10) b = 6.23 and log_(10) c = 7.89`. Find the number of digits before dicimal in ` (ab^(2))/c`. |
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Answer» Correct Answer - 11 `N = (ab^(2))/c` `:. Log_(10) N = log_(10) a+ 2log_(10) b - log_(10) c = 10.28` So, characteristic of N is 10`. So, number of digits before decimal is 11. |
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| 94. |
The `x , y , z`are positive real numbers such that `(log)_(2x)z=3,(log)_(5y)z=6,a n d(log)_(x y)z=2/3,`then the value of `(1/(2z))`is ............ |
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Answer» Correct Answer - 5 ` z = 8x^(3), z = 5^(6)y^(6), z = x^(2//3) y^(2//3)` ` rArr x = (z^(1//3))/2, y = (z^(1//6))/5` ` rArr z = 1/(2^(2//3)) z^(2//9) * (z^(2//18))/(5^(2//3))` ` rArr z^(2//3) = 1/((10)^(2//3)) or z = 1/(10)` |
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| 95. |
If x and y are positive real numbers such that ` 2log(2y - 3x) = log x + log y," then find the value of " x/y`. |
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Answer» Correct Answer - `4/9` We have ` 2log(2y - 3x)= log x + log y` ` rArr log(2y-3x)^(2) = log xy` ` rArr 4y^(2) - 12xy + 9x^(2) = xy` `rArr 4y^(2) - 13xy + 9x^(2) = 0` ` rArr (4y-9x)(x-y) = 0` ` rArr 4y = 9x ` (As for x = y , LHS is not defined) ` rArr x/y = 4/9` |
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| 96. |
If `xa n dy`are real numbers such that `2log(2y-3x)=logx+logy` ,then find `x/y`. |
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Answer» `2log(2y-3x)=logx+logy` `log(2y-3x)^2=logxy` `(2y-3x)^2=xy` `y^2[2-3x/y]^2=xy` `4+9(x/y)^2-12x/y=x/y` Let `x/y=t` `4+9t^2-12t-t=0` `9t^2-13t+4=0` `9t^2-9t-4t+4=0` `9t(t-1)-4(t-1)=0` `(t-1)(9t-4)=0` `t=1,4/9` `x/y=1,4/9`. |
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| 97. |
If `(logx)/(b-c)=(logy)/(c-a)=(logz)/(a-b)`, then which of the following is/are true?`z y z=1`(b) `x^a y^b z^c=1``x^(b+c)y^(c+b)=1`(d) `x y z=x^a y^b z^c`A. ` xyz = 1`B. ` x^(a)y^(b)z^(c) = 1`C. ` x^(b+c) y^(c+a) z^(a+b) = 1`D. ` xyz = x^(a) y^(b) z^(c)` |
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Answer» Correct Answer - A::B::C::D Let `(log_(k)x)/(b - c) = (log_(k)y)/(c-a) = (log_(k) z)/(a-b) = p` ` rArr x = k^(p(b-c)),y=k^(p(c-a)),z = k^(p(a-b))` ` rArrxyz = k^(p(b-c))k^(p(c-a))k^(p(a-b))` ` = k^(p(b-c)+p(c-a)+p(a-b))=k^(0) = 1` `x^(a)y^(b)z^(c) = k^(pa(b-c))k^(pb(c-a))k^(pc(a-b))=k^(0)=1` ` x^(b+c)y^(c+a)z^(a+b) = k^(p(b+c)(b-c))k^(p(c+a)(c-a))k^(p(a+b)(a-b))` ` =k^(0) = 1` |
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| 98. |
If `a^2+ b^2=7ab`, prove that `log(1/3 (a+ b)) =1/2 (log a+ log b)` |
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Answer» `a^2+b^2=7ab` adding 2ab on both sides `a^2+b^2+2ab=7ab+2ab` `(a+b)^2=9ab` `(a+b)=3sqrt(ab)` `(a+b)/3=(ab)^(1/2)` taking log both sides `log((a+b)/3)=log(ab)^(1/2)` `log;1/3(a+b)]=1/2logab` `log[1/3(a+b)]=1/2[loga+logb]`. |
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| 99. |
Prove that `(2^(log_(2^(1//4))x)-3^(log_(27)(x^(2)+1)^(3))-2x)/(7^(4log_(49)x)-x-1)gt0,AAx in(0,oo)`. |
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Answer» `(2^(log2^(1//4)x)-3^(log_(27)(x^(2)+1)^(3))-2x)/(7^(4log_(49)x)-x-1)` `=(2^(log_(2)x^(4))-3^(log_(3)(x^(2)+1))-2x)/(7^(log_(7)x^(2))-x-1)` `(x^(2)-(x^(2)+2x+1))/(x^(2)-x-1)` `=x^(2)+x+1` `(x+(1)/(2))^(2)+(3)/(4)gt0,AAx inR` |
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| 100. |
Solve: `(log)_(0. 5)(3-x)/(x+2) |
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Answer» ` log_(0.5).(3-x)/(x+2) lt 0` ` or (3-x)/(x+2) gt (0.5)^(0)` ` or (3-x)/(x+2) gt 1 ` ` or (3-x)/(x+2 ) - 1 gt 0` ` or (3-x-x-2)/(x+2) gt 0` ` or (2x-1)/(x+2) lt 0` ` or -2 lt xlt 1//2` |
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