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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.
| 101. |
Solve ` log_(2)(x-1) gt 4`. |
| Answer» ` log_(2)(x-1) gt 4 or x - 1 gt 2^(4) or x gt 17` | |
| 102. |
Solve `log_(2).(x-4)/(2x+5) lt 1`. |
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Answer» Correct Answer - `x in (- infty, - 14/3)cup (4, infty)` `log_(2).(x-4)/(2x+5) lt 1` ` rArr 0 lt (x-4)/(2x+5) lt 2^(1)` ` rArr (x-4)/(2x+5) gt 0 and (x-4)/(2x+5) lt 2` ` rArr(x-4)/(2x+5) gt 0 and (x-4)/(2x+5) -2 lt 0` `rArr (x-4)/(2x+5) gt 0and (-3x-14)/(2x+5) lt 0` `rArr (x-4)/(2x+5) gt 0 and (3x+14)/(2x+5) gt 0` `rArr x lt - 5//2 or x gt 4 and x lt - 14//3 or x gt - 5//2` ` rArr x in (-infty, -14//3) cup(4, infty)` |
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| 103. |
Solve ` log_(2).(x-1)/(x-2) gt 0 ` . |
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Answer» ` log_(2).(x-1)/(x-2) gt 0 or (x-1)/(x-2) gt 2^(0)` ` or (x-1)/(x-2) gt 1 or (x-1)/(x-2) - gt 0` ` or (x-1-x+2)/(x-2) gt 0` ` or 1/(x-2) gt 0 or x gt 2` |
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| 104. |
Solve `|x-1|^((log_(10) x)^2-log_(10) x^2=|x-1|^3` |
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Answer» We have `|x-1|^((log_(10)x)^(2)-log_(10)x^(2))=|x-1|^(3)`. `rArr x = 2 or (log_(10)x)^(2)-2log_(10)x-3=0` ` rArr x = 2 or (log_(10)x-3)(log_(10)x+1)=0` ` rArr x = 2 or log_(10)x = 3, log_(10) x =- 1` ` rArr x = 2, x= 1000 or x = 0.11` |
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| 105. |
Solve `log_(3)|x|gt 2`. |
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Answer» Correct Answer - `x lt -9 or x gt 9` ` log_(3)|x| gt 2` ` or |x| gt 3^(2)` ` rArr x lt -9 or x gt 9` |
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| 106. |
If `a=(log)_(12)18 , b=(log)_(24)54 ,`then find the value of `a b+5(a-b)dot` |
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Answer» `a=(log_2 18)/(log_2 12)=(1+2log_2^3)/(2+log_2^3)-(1)` `b=log_24^54` `b=(log_2^54)/(log_2^24)=(1+3log_2^3)/(3+log_2^3)-(2)` From equation 1 `2a+at=1+2t` `2a-1=(2-a)t` `t=(2a-1)/(2-a)-(a)` `3a+bt=1+3t` `3b-1=t(3-b)` `t=(3b-1)/(3-b)-(b)` From a and b `(2a-1)/(2-a)=(3b-1)/(3-b)` `6a-2ab-33+b=6b-2-3ab+b` `6a-2ab+b-6b+3ab-a=3-2` `ab+5a-5b=1` `ab+5(a-b)=1`. |
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| 107. |
The equation `x^((3/4)(log_2 x)^2 + (log_2 x) - (5/4))=sqrt(2)`has`(1)`at least one real solution`(2)`exactly three solutions`(3)`exactly one irrational solution`(4)`complex roots |
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Answer» `x^((3/4)(log_2x)^2 +log_2x - 5/4) = sqrt2` Taking logs both sides, `=>((3/4)(log_2x)^2 +log_2x - 5/4)log_2 x = log_2 2^(1/2)` `=>((3/4)(log_2x)^2 +log_2x - 5/4)log_2 x = 1/2` Let `log_2x = t` Then, equation becomes, `=>(3/4t^2+t-5/4)t = 1/2` `=>3t^3+4t^2-5t-2 = 0` `=>(t-1)(3t^2+7t+2) = 0` `=>(t-1)(3t^2+6t+t+2) = 0` `=>(t-1)(t+2)(3t+1) = 0` `=>t = 1 or t = -2 or t = -1/3` `=>log_2x = 1 or log_2x = -2 or log_2x = -1/3` `=>x = 2^1,2^(-1/3) , 2^(-2)` So, there are exactly three solutions for `x`. |
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| 108. |
If `(log)_k xdot(log)_5k=(log)_x5,k!=1,k >0,`then `x`is equal to`k`(b) 1/5(c) 5 (d) none of these |
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Answer» `log_y^x=log_c^x/log_c^y` `(logx/logk)*(logk/log5)=log5/logx` `(logx)^2=(log5)^2` `logx=log5` `x=5`. |
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| 109. |
Prove that the equation `x^(log_(sqrtx^(2x)))=4` has no solution. |
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Answer» We have ` x^(log_(sqrtx^(2x)))=4, x gt 0 , x ne 1` ` rArr x ^(2 log_(x) 2x) = 4` ` rArr x^(log_(x)4x^(2))=4` ` rArr 4x^(2) = 4` ` rArr x = pm 1`,which is not possible. Hence, equation has no solution. |
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| 110. |
Solve `1/4 x ^(log_(2)sqrtx)=(2*x^(log_(2)x))^(1/4).` |
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Answer» Taking log on both sides with base 2, we get ` log_(2)(1/4)+(log_(2)sqrtx)(log_(2)x)=1/4+1/4(log_(2)x)^(2)` ` or (log_(2)x)^(2) = 9 or log_(2)x = pm3` ` :. x = 8 or 1/8` |
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| 111. |
`3^(log_3log sqrtx) -log x + (log x)^2 - 3 =0` |
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Answer» `3^(log_3 logsqrtx) - logx +(logx)^2 - 3 = 0` We know, `a^(log_a b) = b`, so our equation becomes, `=>logsqrtx - logx +(logx)^2 - 3 = 0` `=>1/2logx - logx +(logx)^2 - 3 = 0` `=>(logx)^2 - logx/2 - 3 = 0` `=>2(logx)^2 - logx - 6 = 0` `=>2(logx)^2 - 4logx+3logx - 6 = 0` `=>(2logx+3)(logx - 2) = 0` `=>logx = -3/2 and logx = 2` `=>x = e^(-3/2) and x = e^2.` |
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| 112. |
If P is the number of natural numbers whose logarithms to the base 10 have the the charecteristic p and Q is the numbers of natural numbers logarithms of whose reciprocal to the base 10 have the charecteristics -q. then find the value of `log_(10) P-log_(10) Q` |
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Answer» `10^p<=p<10^(p+1)` `p=10^(p+1)-10^p` `=10^p*q` `=9.10^p` Simialrly, `10^(q-1)ltQ<=10^q` `Q=10^q-10^(q-1)` `=9*10^(q-1)` `log_10^p-log_10^Q=log_10^(p/q)` `log_10((9*10^p)/(9*10^(q-1)))` `log_10 10^(p-q+1)=p-q+1`. |
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| 113. |
If `p , q in N`satisfy the equation `x^sqrt(x)=(sqrt(x))^x,`then `pa n dq`are(a)relatively prime (b) twin prime (c)coprime(d)if `(log)_q p`is defined, then `(log)_p q`is not and vice versa |
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Answer» `x^sqrtx=sqrtx^x` taking log both side `sqrtxlogx=xlogsqrtx` `sqrtxlogx=x/2logx` `sqrtxlogx-x/2logx=0` `logx[sqrtx-x/2]=0` either log x=0 `x=1` `sqrtx=x/2` `2sqrtx=x` `4x=x^2` `x=4` `p=1` `q=4` Option a,c,d are correct. |
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| 114. |
Using logarithms, find the value of 6.45 x 981.4 |
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Answer» Let x=6.45*981.4 taking log both side logx=log6.45+log981.4 logx=0.8096+2.9919 logx=3.8015 x=antilog(3.8015) x=6331. |
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| 115. |
If `p,q in N" satisfy the equation "x^(sqrtx) = (sqrtx)^(x)`, thenA. p+q=5B. `|p-q|=4`C. pq=4D. if ` log_(q) p ` is defined, then ` log_(p)q ` is not and vice versa |
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Answer» Correct Answer - A::C::D ` x ^(sqrtx)=(sqrtx)^(x) ,p,qinN` `rArr sqrtx log x = x log sqrtx` ` or log x [sqrtx - x/2] = 0` ` rArr log x = 0 or [sqrtx - x/2] = 0` ` rArr x = 1 or 4` |
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| 116. |
If `(log)_a x=b`for permissible values of `aa n dx ,`then identify the statement(s) which can be correct.If `aa n db`are two irrational numbers, then`x`can be rational.If `a`is rational and `b`is irrational, then `x`can be rational.If `a`is irrational and `b`is rational, then `x`can be rational.If `aa n d b`are rational, then `x`can be rational.A. If a and b are two irrational numbers, then x can be retional.B. If a is rational and b is irrational, then x can be rational.C. If a is irrational and b is rational, then x can be rational.D. If a and b are rational, then x can be rational. |
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Answer» Correct Answer - A::B::C::D ` log_(a) x = b or x = a^(b)` (1) For `a=sqrt2^(sqrt2)!inQ and b = sqrt2 !in Q, x = (sqrt2^(sqrt2))^(sqrt2)` which is rational. (2) For ` a = 2 in Q abd b = log_(2) 3 !in Q , x = 3` which is rational. (3) For ` a = sqrt2 and b = 2 , x = 2` (4) The option is aboviously correct. |
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| 117. |
If`((log)_a N)/((log)_c N)=((log)_a N-(log)_b N)/((log)_b N-(log)_c N),w h e r eN >0a n dN!=1, a , b , c >0`and not equal to 1, then prove that `b^2=a c` |
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Answer» `log_a^N/log_c^N=(log_a^N-log_b^N)/(log_a^N-log_c^N)` `(logN/loga)/(logN/logc)=((logN/loga)-(logN/logb))/((logN/logb)-(logN/logc))` `logc/loga=((logb-loga)/(logalogb))/((logc-logb)/(logblogc))` `logc/loga=((logb-loga)logc)/((logc-loga)loga)` `logc-logb=logb-loga` `loga+logc=2logb` `logac=logb^2` `b^2=ac`. |
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| 118. |
If `(log)_3{5+4(log)_3(x-1)}=2,`then `x`is equal to4 (b) 3(c) 8 (d) `(log)_2 16`A. 2B. 4C. 8D. 16 |
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Answer» Correct Answer - B We must have ` x - 1 gt 0` ` rArr x gt 1`…(i) ` and 5+4 log_(3) (x-1) gt 0` ` or 4 log_(3) (x-1) gt - 5` ` or log_(3) (x-1) gt - 5/4` ` or x - 1 gt 3^(-5//4)` ` or x gt 1+3^(-5//4)` …(ii) From Eqs. (i) and (ii), we get ` x gt 1 + 3^(-5//4)` `:. 5+4 log_(3) (x-1) = 9` ` rArr 4 log_(3) (x-1) = 4` ` or log_(3) (x-1) = 1` ` or x -1 = 3 or x = 4` |
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| 119. |
If `P`is the number of natural numbers whose logarithms to the base 10 havethe characteristic `pa n dQ`is the number of natural numbers logarithms of whose reciprocals to thebase 10 have the characteristic `-q`, then find the value of `log_(10)P-(log)_(10)Qdot` |
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Answer» `10^(p) leP lt 10^(p+1)` ` rArr P=10^(p+1) - 10^(p)` ` rArr P=9 xx 10^(p)` Similarly, `10^(q-1)lt Q le10^(q)` `rArr Q = 10^(q)-10^(q-1) = 10^(q-1)(10-1)=9 xx 10^(q-1)` ` :. log_(10)P-log_(10)Q = log_(10)(P//Q)` ` log_(10)10^(p-q+1)` ` p-q+1` |
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| 120. |
`(log)_4 18`isa rational number(b) an irrational numbera prime number(d) none of these |
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Answer» `=log_4 18` `=log_4(9*2)=log_4 9+log_4 2` `=log_(2^2)^(3^2)+log_(2^2)^2` `=(2/2)log_2^3+1/2log_2^2` `=log_2^3+1/2` `=1+1/2=3/2` This is rational. |
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| 121. |
Which of the following, when simplified, reduces to unity?`(log)_(10)5dot(log)_(10)20+((log)_(10)2)^2`. (c) `-(log)_5(log)_3sqrt(5sqrt(9))``1/6(log)_((sqrt(3))/2)((64)/(27))`A. ` log_(10) 5 * log_(10) 20+(log_(10) 2)^(2)`B. `(2 log 2+log 3)/(log 48 - log 4)`C. ` - log_(5) log_(3) sqrtroot(5)9`D. `1/6 log_(sqrt3//2)(64/27)` |
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Answer» Correct Answer - A::B::C (1) ` log_(10)(10/2) * log_(10)(10xx2)+(log_(10)2)^(2)` ` = (1-log_(10)2)(1+log_(10)2)+(log_(10)2)^(2)=1` (2) `(log2^(2)xx3)/(log(48//4)) = 1` (3) ` -log_(5)log_(3)9^(1//10)=- log_(5)log_(3)3^(1//5)` ` = - log_(5)(1//5)=1` (4) ` 1/4 log_(sqrt3/2)(64/27) = 1/6 log_(sqrt3/2)(sqrt3/2)^(-6)=- 1` |
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| 122. |
The value of `b`for which the equation `2(log)_(1/(25))(b x+28)=-(log)_5(12-4x-x^2)`has coincident roots is`b=-12`(b) `b=4orb=-12``b=4orb=-12`(d)`b=-4orb=12` |
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Answer» LHS `2log_(1/25)(bx+28)2log_(5^(-2))(bx+28)` LHS `(2/-2)log_5(bx+28)=-log_5(bx+28)` `log_5(bx+28)=-log_5(12-4x-x^2)` `log_a^x=log_a^y` `x=y` `bx+28=12-4x-x^2` `x^2+4x+bx+28-12=0` `x^2+(4+b)x+16=0` `(4+b)^2-4*16=0` `16+b^2+8b-64=0` `b^2+8b-48=0` `(b+12)(b-4)=0` `b=4,-12`. |
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| 123. |
If `A=log_2log_2log_4(256)+2log_(sqrt(2))(2)`, then find the value of A. |
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Answer» `A=log_2log_2log_4^(256)+2log_sqrt2^2` `A=log_2log_2^4+2/sqrt2log_2^2` `A=log_2log_2^(2^2)+4log_2^2` `A=log_2^2+4` `A=1+4=5`. |
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| 124. |
Solution set of the inequality `1/(2^(x) - 1) gt 1/(1-2^(x-1))` isA. ` (1, infty)`B. ` (0, log_(2) (4//3))`C. ` (-1, infty)`D. ` (0, log_(2)(4//3))cup(1, infty)` |
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Answer» Correct Answer - D Put ` 2^(x) = t." Then " t gt 0`. The given inequality becomes ` 1/(t-1) gt 2/(2 - t)` ` rArr 1/(t-1) - 2/(2 - t) gt 0` ` rArr (2-t - 2t + 2)/((t-1)(2-t)) gt 0` ` rArr (4 - 3t)/((t-1)(2-t)) gt 0` ` rArr (t-1)(t-4//3)(t-2) gt 0` . From above sign scheme, we get ` 1 lt t 4//3 or t gt 2`. `rArr 1 lt 2^(x) lt 4//3 or 2^(x) gt 2` ` rArr 0 lt x lt log_(2) (4//3) or x gt 1` |
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| 125. |
Which of the following is not the solution of `(log)_3(x^2-2) |
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Answer» Correct Answer - C We have, ` log_(3) (x^(2) - 2) lt log_(3) (3/2|x|-1)` For this to be true, we must have `x^(2) - 2 gt 0, 3/2 |x| - 1 gt 0` ` and x^(2) - 2 lt 3/2 |x|-1` ` rArr x ^(2) gt 2, |x|gt 2/3 and 2|x|^(2) - 3 |x|-2 lt 0` ` rArr |x| gt sqrt2, |x| gt 2/3 and (2|x|+ 1)(|x|-2) lt 0` ` rArr |x| gt sqrt2 and |x| lt 2` ` rArr sqrt2 lt |x| lt 2` ` rArr x in (-2, - sqrt2) cup(sqrt2, 2)` |
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| 126. |
The number of `N=6-(6(log)_(10)2+(log)_(10)31)`lies between two successive integers whose sum is equal to(a)5 (b) 7 (c) 9 (c) 10 |
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Answer» `N in(I_1,I_2)` `N=6-[6log_10^2+log_10^31]` `N=6-[log_10^(2^6)+log_10^31]` `N=6-[log_10^32+log_10^31]` `N=6-[log_10^(32*31)]` `N=6-log_10^992` `N=6-log_10^(10^3-8)]` `N=6-(3^-)` `N=6-(3-h)` `N=6-3+h` `N=(3+h)` Sum=3+4=7. |
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| 127. |
Write the significant digits in each of the following numbers tocompute the mantissa of their logarithms:3.239 (ii) 8(iii) 0.9(iv) 0.020.0367 (vi) 89 (vii)0.0003 (viii) 0.00075 |
| Answer» `{:("Number",,"Significant digits"),(3.239,," "3239),(8,," "80),(0.9,," "90),(0.02,," "20),(0.0367,," "367),(89,," "89),(0.0003,," "30),(0.00075,," "75):}` | |
| 128. |
Find the antilogarithm of each of the following:2.7523 (ii) 3.7523 (iii) 5.7523 (iv) 0.75231.7523 (vi) 2.7523 (vii) 3.7523 |
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Answer» 1)antilogarithm of 2.7523 is 565.3 2)antilogarithm of 3.7523 is 5655.0 3)antilogarithm of 5.7523 is 565300.0 4)antilogarithm of 0.7523 is 5.653 5)antilogarithm of 1.7523 is 0.5653 6)antilogarithm of 2.7523 is 0.5653 7)antilogarithm of 3.7523 is 0.005653. |
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| 129. |
`(log sqrt(27) +log8 -log sqrt(1000))/ log 1.2 ` Find the value |
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Answer» We know `log_ab^c=clog_ab` using this property our expression simplifies to =>`(3/2log(3)+3log2-3/2)/(log4+log3-1)` => taking `3/2` common we get=>`3/2``((log4+log3-1)/(log4+log3-1))`=> final ans=`3/2`. |
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| 130. |
Find the value of : (i) antilog 0.654 (ii) antilog 1.204 (iii) antilog `bar(1).3612` (iv) antilog `bar(3).4568` |
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Answer» We have (i) antilog (0.654) = 4.508 (ii) antilog (1.204) = 16.00 (iii) antilog `(bar(1).3612) = 0.2297` (iv) antilog `(bar(3).4568) = 0.002863` |
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| 131. |
If `S={x in R :((log)_(0. 6)0. 216)(log)_5(5-2x)lt=0},`then `S`is equal to`(2. 5 ,oo)`(b) `(2, 2.5)`(c) `(2, 2.5)`(d) `(0, 2.5)`A. `[ 2.5, infty)`B. ` [2, 2.5)`C. ` (2, 2.5)`D. `(0, 2.5)` |
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Answer» Correct Answer - B ` ( log_((0.6))(0.6)^(3)) log_(5) (5-2x) le 0` ` rArr 5 - 2x le 1 ` ` rArr x ge 2 ` ...(i) Also`, 5 - 2x gt 0` ...(ii) From Eqs. (i) and (ii), we have ` x in [2, 2.5)`. |
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| 132. |
`x^((log)_5x)>5`implies`x in (0,oo)`(b) [2,2.5] (c)(2,2.5) (d) (0,2.5)A. ` x in (0, infty)`B. ` x in (0, 1//5) cup (5, infty)`C. ` x in (1, infty)`D. ` x in (1, 2)` |
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Answer» Correct Answer - B `x^(log_(5) x) gt 5` Taking logarithm with base 5, we have `(log_(5)x) (log_(5)x) gt 1` ` or (log_(5)x-1) (log_(5)x+1) gt 0` ` or log_(5) x gt 1 or log_(5) x lt - 1` `rArr x gt 5 or x lt 1//5` Also we must have ` x gt 0` . Thus, ` x in (0, 1//5) cup (5, infty)`. |
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| 133. |
Solution set of the inequality `(log)_(0. 8)((log)_6(x^2+x)/(x+4)) |
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Answer» Correct Answer - D ` log_(0.8)(log_(6).(x^(2)+x)/(x+4)) lt 0` ` rArr log_(6). (x^(2)+x)/(x+4) gt 1 ` ` rArr (x^(2)+x)/(x+4) gt 6` ` rArr (x^(2)+x)/(x+4) - 6 gt 0` ` rArr (x^(2) - 5x - 24)/((x+4)) gt 0` ` rArr ((x-8)(x+3))/((x+4)) gt 0` From the sign scheme of ` ((x-8)(x+3))/((x+4))`, ` x in (-4, -3) cup (8, infty)`. |
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| 134. |
Solve `log_(1//2)(x^(2)-6x+12) ge -2`. |
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Answer» Correct Answer - ` x in [2, 4]` `x^(2) - 6x+ 12 le (1/2)^(-2)` ` or x^(2) - 6x + 8 le 0` ` or (x-2)(x-4) le 0` ` rArr x in [2, 4]` `"Also, " x^(2) - 6x+12 gt 0 or (x-3)^(2) + 3 gt 0" which is true for any real x. Hence, "x in [2, 4]`. |
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| 135. |
Given that `log(2)=0. 3010 ,`the number of digits in the number `2000^(2000)`is6601 (b)6602 (c) 6603 (d) 6604 |
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Answer» `log_10^2=0.3010` `2000log2000` `2000(log_10^2+log_10^(10^3))` `2000(0.3010+3)` `2000(3.3010)` `6602.0`. |
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| 136. |
Fine the value of : (i) log 3 (ii) log 23.45 (iii) log 4.17 (iv) log 0.325 (iv) log 0.0954 (vi) log 0.00056 |
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Answer» Using log table, we have (i) log 3 = 0.4771 (ii) log 23.45 = 1.3701 (iii) log 4.17 = 0.6201 (iv) log `0.325 = bar(1).5119` (v) log `0.0954 = bar(2).9795` (vi) log `0.00056 = bar(4).7482` |
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| 137. |
The true solution set of inequality `(log)_((x+1))(x^2-4)>1`is equal to`2,oo)`(b) `(2,(1+sqrt(21))/2)``((1-sqrt(21))/2,(1+sqrt(21))/2)`(d) `((1+sqrt(21))/2,oo)` |
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Answer» `log_(x+1) (x^2-4)>1` `x in(2,oo)` `0ltx+1<1` `-1ltxlt0` this is not possible `x+1>1` `x>0` `x^2-4>x+1` `x^2-x-5>0` `x in((1+sqrt21)/2,oo)`. |
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| 138. |
Which of the following is not the solution `log_(x)(5/2-1/x) gt (5/2-1/x)` ?A. `(2/5,1/2)`B. `(1,2)`C. ` (2/5, 1)`D. none of these |
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Answer» Correct Answer - A :: B Given inequality is defined if ` x gt 2//5, x ne 1` Case I: If ` x gt 1` ` rArr 5/2 - 1/x gt x ` ` or x + 1/x lt 5/2` ` or 2(x^(2) + 1) lt 5x` ` or 2x^(2) - 5x + 2 lt 0 ` ` or 2 x^(2) - 4x - x + 2 lt 0` `or (x-2)(2x-1) lt 0` ` rArr x in (1, 2)` ....(i) Case II: ` 2/5 lt x lt 1, " then " (x-2)(2x-1) gt 0` ` rArr x in (2/5,1/2)` ...(ii) From Eqs. (i) and (ii), we get ` x in (2/5,1/2) cup (1, 2) `. |
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| 139. |
If `a^x=b ,b^y=c ,c^z=a ,`then find the value of `x y zdot` |
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Answer» `a^x=b,b^y=e,c^z=a` taking log both side `xloga=logb` `x=logb/loga` Similarly, `y=logc/logb` `z=loga/logc` `logb/loga*logc/logb*loga/logc=1`. |
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| 140. |
Solve `(x+1)^(log_(10) (x+1))=100(x+1)`A. all the roots are positive real numbers.B. all the roots lie in the interval (0, 100)C. all the roots lie in the interval [-1, 99]D. none of these |
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Answer» Correct Answer - C `(x+1)^(log_(10)(x+1))=100(x+1)` ` rArr log_(10) (x+1)^(log_(10)(x+1))=log_(10) (100(x+1))` ` rArr log_(10)(x+1)log_(10)(x+1) = 2 + log_(10) (x+1)` Let ` log_(10)(x+1) = y` ` rArr y^(2)-y-2=0` ` rArr y = 2 or - 1` ` rArr log_(10)(x+1) = 2, -1` ` rArr x+1 = 100, 1//10` ` rArr x =99 or - 9//10` |
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| 141. |
Solve: `(log)_((2x+3))(6x^2+23+21)+(log)_((3x+7))(4x^2+12 x+9)=4` |
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Answer» The given equation can be written as ` log_((2x+3))(2x+3)(3x+7)+log_((3x+7))(2x+3)^(2)=4` ` or log_((2x+3))(2x+3)+log_((2x+3))(3x+7)` `+2/(log_((2x+3))(3x+7))=4` Let `log_((2x+3))(3x+7)=t`.Then ` 1+t+2/t = 4` ` or t^(2) = 3t+2 = 0` ` or (t-1)(t-2) = 0` `rArr t=1, t = 2` (i) if t = 1, then ` log_((2x+3))(3x+7)=1` ` or 3x+7 = 2x + 3` Hence, x =- 4, which is not possible as ` 2x+3 gt 0 and 3x + 7 gt 0`. (ii) if t = 2, then ` log_((2x+3))(3x+7)= 2` ` or 3x+7 = (2x+3)^(2)` ` or 4x^(2) + 9x + 2 = 0` ` or (x+2) (4x+1) = 0` ` rArr x =- 2 and x =- 1/4` Sincex =- 2 is not possible, there is only one solution, ` x =- 1//4`. |
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| 142. |
Q.15 Simplify: log 2+log +log 1/2 5 us 10- 2 21 |
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Answer» `5^(log_(1/5)(1/2)) = 5^(log_(5^(-1))(2^(-1))` `= 5^(log_(5)(2)) = 2` Now, `log_(sqrt2) (4/(sqrt7+sqrt3)) = log_(2^(1/2)) (4/(sqrt7+sqrt3)**(sqrt7- sqrt3)/(sqrt7- sqrt3)) ` `=2log_2 ((4(sqrt7-sqrt3))/(7-3))` `=log_2 (sqrt7- sqrt3)^2` `=log_2 (10-2sqrt21**(10+2sqrt21)/(10-2sqrt21))` `=log_2 ((100-84)/(10-2sqrt21))` `=log_2 (16/(10-2sqrt21))` `=>log_2 16 + log_2 (1/(10-2sqrt21))` Now, the given equation becomes, `5^(log_(1/5)(1/2))+log_(sqrt2) (4/(sqrt7+sqrt3))+ log_(1/2) (1/(10-2sqrt21))) = 2+log_2 16 + log_2 (1/(10-2sqrt21)) + log_(1/2) (1/(10-2sqrt21))` `2+log_2 16 + log_2 (1/(10-2sqrt21)) - log_2 (1/(10-2sqrt21))` `2+4`...[As `log_2 16 = 4`] `=6.` `:. 5^(log_(1/5)(1/2))+log_(sqrt2) (4/(sqrt7+sqrt3))+ log_(1/2) (1/(10-2sqrt21))) = 6` |
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| 143. |
Sum of integers satisfying ` sqrt(log_(2)x-1)-1//2 log_(2)(x^(3))+2 gt 0` is ________. |
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Answer» Correct Answer - 5 `sqrt(log_(2) x-1)-3/2 log_(2) x+2 gt 0 (xgt 0)` ` rArr sqrt(log_(2)x-1)-3/2 (log_(2) x-1)+1/2 gt 0` ...(i) Let ` sqrt(log_(2)x-1)=tge 0`, we have ….(ii) ` log_(2) x-1 ge 0` Then from Eq. (i), we have ` t-3/2 t^(2) + 1/2 gt 0` ` 3t^(2) - 2t -1 lt 0 ` ` rArr 1//3 lt t lt 1` ...(iii) Form Eqs. (ii) and (iii), we have ` 0 le t lt 1`. ` 0 le sqrt(log_(2)x-1) lt 1` ` 0 le log_(2) x - 1 lt 1` ` 1 le log_(2) x lt 2` ` 2 le x lt 4` Hence, the integral values are 2 and 3, and their sum is 5. |
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| 144. |
If `(log)_3y=xa n d(log)_2z=x ,`find `72^x`in terms of `yand zdot` |
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Answer» `(72)^x` `(2^3*3^2)^x` `2^(3x)*3^(2x)` `(2^x)^3*(3^x)^2` `z^3*y^2` `y^2z^3`. |
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| 145. |
Evaluate `(72.3)^(1/3); if `log 0.723=bar(1).8591` |
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Answer» Let `root(3)72.3`. Then ` log x = log(72.3)^(1//3)` ` or log x = 1/3 log 72.3` ` or log x = 1/3 xx 1.8591` ` rArr log x = 0.6197` ` or x = "antilog "` (0.6197)` ` = 4.166 ` (using antilog table) |
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| 146. |
If `x=log_(2a) a,y=log_(3a) 2a ` and `z=log_(4a) 3a` then prove that `xyz+1=2yz` |
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Answer» `1+xyz=1(log_(2a)a)(log_(3a)2a)(log_(4a)3a)` `=1+(loga)/(log2a)(log2a)/(log3a)(log3a)/(log4a)` `=1+(loga)/(log4a)` `=log_(4a)4a+log_(4a)a` `=log_(4a)4a^(2)=2log_(4a)2a` `=2(log_(3a)2a)(log_(4a)3a)=2yz` |
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| 147. |
For `a >0,!=1,`the roots of the equation `(log)_(a x)a+(log)_x a^2+(log)_(a^2a)a^3=0`are given`a^(-4/3)`(b) `a^(-3/4)`(c) `a`(d) `a^(-1/2)`A. `a^(-4//3)`B. `a^(-3//4)`C. aD. `a^(-1//2)` |
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Answer» Correct Answer - A::D ` log_(ax)a+log_(x)a^(2)+log_(a^(2)x)a^(3) = 0` ` or 1/(log_(a)ax)+2/(log_(a)x)+3/((log_(a)a^(2)x))= 0` `or 1/(1+log_(a)x) + 2/(log_(a) x) + 3/((2+log_(a)x )) = 0` Let ` log_(a)x = y ," we have" 1/(y+1) + 2/y+3/(2+y) = 0` ` or 6y^(2) + 11y+4 = 0` ` or y = log_(a) x = - 1/2, - 4/3` ` rArr x = a^(-4//3), a^(-1//2)` |
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| 148. |
Solve `log_4 (8)+log_4 (x+3)-log_4 (x-1)=2` |
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Answer» `log_(4)^(8)+log_(4)(x+3)-log_(4)(x-1)=2` `or"log"_(4) (8(x+3))/(x-1)=2` or `(8(x+3))/(x-1)=4^(2)` `orx+3=2x-2` `rArr" "x=5` Also for x=5 all terms of the equation are defined. |
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| 149. |
Solve `x^((log)_y x)=2a n dy^((log)_x y)=16`A. ` 2^(root(3)2)`B. ` 2 ^(root(3)4)`C. ` 2 ^(root(3)128)`D. ` 2 ^(root(3)16)` |
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Answer» Correct Answer - D Let ` log_(y) x = 1` Then ` x = y^(t)` …(1) Now, ` x^(log_(y) x) =2` becomes ` x^(t) = 2` `rArr x = 2^(1//t)` …(2) And `y^(log_(x)y) = 16` becomes ` y^(1//t) = 2^(4)` ` rArr y = 2^(4//t)` ….(3) Putting the values of x and y in (1), we get ` 2^(1//t) = 2^(4t^(2))` ` rArr 4t^(3) = 1` ` :. t = (1/4)^(1//3)` ....(4) Using (4) and (2), we get ` x = (2)^((4)^(1//3)) = 2^(root(3)4)` Using (4) and (3), we get ` y = (2)^((4)^(2//3)) = 2^(root(3)16)` |
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| 150. |
If `(21. 4)^a=(0. 00214)^b=100`, then the value of `1/a-1/b`is0 (b) 1 (c) 2(d) 4 |
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Answer» Correct Answer - C `(21.4)^(a) = 100` ` rArr a log(21.4) = 2` ` :. Log(21.4) = 2//a` …(i) From ` (0.00214)^(b) = 100`, we get ` b(log 0.00214)= 2` ` or b log(214.4 xx 10^(-4)) = 2` ` or b = 2/(log 21.4-4) = 2/(2/a-4) = 1/(1/a - 2) = a/(1-2a)` ` rArr 1/b = (1-2a)/a` ` rArr 1/a - 1/b = 2` |
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