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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. |
Solve `log_(2)(4xx3^(x)-6)-log_(2)(9^(x)-6)=1`. |
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Answer» `log_(2)(4xx3^(x)-6)-log_(2)(9^(x)-6)=1` or `"log"_(2)(4xx3^(x)-6)/(9^(x)-6)=1` or `"log"_(2)(4xx3^(x)-6)/(9^(x)-6)=2` `or4y-6=2y^(2)-12" "("putting"3^(x)=y)` `ory^(2)-2y-3=0` `ory=-1,3` `or3^(x)=3` `orx=1` |
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| 2. |
Let a, b, c, d be positive integers such that ` log_(a) b = 3//2 and log_(c) d = 5//4`. If (a-c) = 9, then find the value of (b-d). |
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Answer» `b = a^(3//2) and d=c^(5//4)` Let ` a= x^(2) and c=y^(4), x, y in N` ` rArr b= x^(3), d= y^(5)` Given ` a-c=9, " then "x^(2)-y^(4) = 9` ` rArr (x- y^(2))(x+y^(2)) = 9` . Hence, ` x- y^(2) = 1 and x + y^(2) =9`. (No other combination in the set of positive integers will be possible.) ` x= 5 and y = 2` `:. b - d = x^(3) - y^(5) = 125 - 32 = 93` |
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| 3. |
If ` log_(b) n = 2 and log_(n) 2b = 2`, then find the value of b. |
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Answer» Correct Answer - `2^(1//3)` Eliminating n, we have ` log_(b) 2b = 4` ` or 2b = b^(4)` ` or b^(3) = 2` ` rArr b = 2^(1//3)` |
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| 4. |
If ` log_(2) x xx log_(3) x = log_(2) x + log_(3) x`, then find x . |
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Answer» Correct Answer - x = 1, 6 ` log_(2) x xx log_(3) x = log_(2) x + log_(3) x` ` rArr (log x)/(log 2) *(log x)/(log 3) =(log x)/(log 2) +(log x)/(log 3) ` ` rArr log x = 0` ` or (log x)/(log 2*log 3) = 1/(log 2) + 1/(log 3) ` ` rArr x = 1 or log x = log 2+ log 3` `rArr x = 1 or log x = log 6` ` rArr x = 1 or x = 6` |
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| 5. |
If `y^2=xz` and a^x=b^y=c^z`then prove that `log_b (a)= log_c (b) |
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Answer» `a^(x) = b^(y) = c^(z)` ` rArr x log a = y log b = z log c` ` :. y/x = z/y rArr(log a)/(log b) = (log b)/(log c)` ` rArr log_(b) a = log_(c) b` |
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| 6. |
Let `a= log_(3) 20, b = log_(4) 15 and c = log_(5) 12`. Then find the value of `1/(a+1)+1/(b+1)+1/(c+1)`. |
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Answer» We have `a+1 = log_(3) 20 + log_(3) 3 = log_(3) 60` ` b+ 1 = log_(4) 15 + log_(4) 4 = log_(4) 60` ` c+1 = log_(5) 12+ log_(5) 5 = log_(5) 60` `1/(a+1)+1/(b+1)+1/(c+1)=1/(log_(3) 60)+1/(log_(4)60)+1/(log_(5)60)` ` =log_(60)3+log_(60)4+log_(60)5` ` = log_(60)(3 xx 4 xx 5)` ` = log_(60) 60` ` = 1` |
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| 7. |
If `a^4b^5=1` then the value of `log_a(a^5b^4)` equalsA. `9//5`B. 4C. 5D. `8//5` |
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Answer» Correct Answer - A Given ` 4 log_(a) a+5log_(a) b =0` ` rArr log_(a) b = -4//5` …(i) Now ` log_(a)(a^(5)b^(4)) = 5+4 log_(a) b = 5 + 4 (-4/5)` ` = 5 - 16/5 = 9/5` |
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| 8. |
`2^((sqrt(log_a(ab)^(1/4)+log_b(ab)^(1/4))-sqrt(log_a(b/a)^(1/4)+log_b(a/b)^(1/4)))sqrt(log_a(b))` =A. 1B. 2C. ` 2^(log_(a) b)`D. ` 2 ^(log_(b)a)` |
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Answer» Correct Answer - C We have ` E= 2^((sqrt(log_(a)root(4)(ab)+log_(b)root(4)(ab))-sqrt(log _(a)root(4)(b/a+log_(b)root(4)(a/b)))) sqrt(log_(a)b))` ` = 2^(1/2(sqrt(log_(a)ab+log_(b)ab-)sqrt(log_(a)b//a+log_(b)a//b))sqrt(log_(a)b))` ` = 2^(1/2(sqrt(2+log_(a)b+log_(b)a)-sqrt(log_(a)b+log_(b)a - 2))sqrt(log_(a)b))` ` = 2^(1/2(sqrt((log_(a)b)^(2)+2log_(a)b+1)-sqrt((log_(a)b)^(2)-2log_(a)b+1))` ` = 2^(1/2(sqrt((log_(a)b+1)^(2))-sqrt((log_(a)b-1)^(2)))` ` =2^(1/2(|log_(a)b+1|-|log_(a)b-1|)` Case I: ` bgea gt1` ` rArr log_(a) b ge log_(a) a` ` rArr log_(a) b ge 1` ` rArrE=2^(1/2(log_(a)b+1-log_(a)b+1))=2` Case II: ` 1 lt b lt a` ` rArr 0 lt log_(a) b lt log_(a) a` ` rArr 0 lt log_(a) b lt 1` ` rArr E = 2^(1/2(log_(a)b+1-1+log_(a)b))` ` = 2^(1//2.(2log_(a)b))` ` = 2 ^(log_(a)b)` |
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| 9. |
Solve for `x: log_(4) log_(3) log_(2) x = 0`. |
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Answer» Correct Answer - x=8 `log_(4)log_(3)log_(2)x = 0` ` rArrlog_(3)log_(2) x = 1` ` rArr log_(2) x = 3` ` rArr x = 2^(3) = 8` |
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| 10. |
Solve ` (1/2)^(x^(6)-2x^(4)) lt 2^((x)^(2))`. |
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Answer» Correct Answer - ` x in R - {0, pm 1}` ` (1/2)^(x^(6)-2x^(4))lt2^(x^(2))` ` or (1/2)^(x^(6)-2x^(4))lt(1/2)^(-x^(2))` ` or x^(6) - 2x^(4) gt -x^(2)` ` or x^(6) - 2x^(4) + x^(2) gt 0` ` or (x^(3)=x)^(2) gt 0` ` rArr x^(3) - x ne 0` ` :. x ne 0, - 1, 1` ` :. x in R - {0, pm 1}` |
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| 11. |
Solve for x and ` y:y^(x) = x^(y), x = 2y`. |
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Answer» Correct Answer - ` x = 4, y = 2` `y^(x) = x^(y), x = 2 y ` ` rArr y^(x) =(2y)^(x//2)` ` rArr y = 2,`hence x = 4 |
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| 12. |
Solve ` 2^(x+2)-2^(x+3) -2^(x+4) gt 5^(x+1) -5^(x+2)`. |
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Answer» Correct Answer - ` x in (0, infty)` `2^(x+2)-2^(x+3) -2^(x+4) gt 5^(x+1) -5 ^(x+2)` ` rArr 2^(x)(4-8-16) gt 5^(x) (5 - 25)` ` rArr (2//5)^(x) lt 1` ` rArr (2//5)^(x) lt (2//5)^(0)` ` rArr x in (0, infty)` |
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| 13. |
Solve` (3/4)^(6x+10-x^(2)) lt 27/64`. |
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Answer» Correct Answer - ` -1 lt x lt 7` `(3/4)^(6x+10-x^(2)) lt27/64` ` rArr(3/4)^(6x+10-x^(2)) lt (3/4)^(3)` `rArr 6x + 10 - x^(2) gt 3` ` rArr x^(2) - 6x - 7 lt 0` ` rArr (x+1)(x-7) lt 0` ` rArr -1 lt x lt 7` |
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| 14. |
Find the number of solutions of ` |x|*3^(|x|) = 1`. |
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Answer» Correct Answer - 2 We have ` |x|3^(|x|)=1` ` or |x| = 3^(-|x|)` Now, ` 3^(-|x|) ={{:(3^(-x)",",x ge 0),(3^(x)" ,",x lt 0):}` To find the number of roots of the above equation, we need to find the number of points of intersection of `y=|x| and y = 3^(|x|)` The graphs of these functions are as shown in the following figure: From the fraph number of solution is 2. |
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| 15. |
Solve ` 2^(log_(2)(x-1))gtx+5`. |
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Answer» Correct Answer - no solution `2^(log_(2)(x-1)) gt x + 5` ` or x - 1 gt x + 5` ` or -1 gt 5 ` which is not possible. |
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| 16. |
Find the value of `3^(2log_(9)3)`. |
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Answer» Correct Answer - 3 ` 3^(2 log_(9)3)=3^(log_(9)9^(1//2)` ` = 3^(2(1/2))=3` |
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| 17. |
Let `f(x) = sqrt(log_(10)x^(2))`.Find the set of all values of x for which f (x) is real. |
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Answer» Correct Answer - `x le - 1 or x ge 1` `log_(10)x^(2) ge 0` ` rArr log_(10)x^(2) ge log_(10) 1` ` rArr x^(2) ge 1` ` rArr x ge 1` ` or x le -1`. |
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| 18. |
Find the values of x which the function `f(x)=sqrt(log_(1//2)((x-1)/(x+5))` is defined. |
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Answer» Correct Answer - ` x in (1, infty)` `f(x)=sqrt(log_(1//2(x-1)/(x+5)))` It is defined if `log_(1//2).(x-1)/(x+5) ge 0` ` rArr 0 lt(x-1)/(x+5) le 1` When `(x-1)/(x+5) gt 0, x in (-infty, -5) uu(1, infty)` ...(i) When `(x-1)/(x+5) le 1` `rArr (x-1)/(x+5) - 1 le 0` `rArr (-6)/(x+5) le 0` ` rArr x gt - 5` ...(ii) From (i) and (ii), ` x in (1, infty)`. |
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| 19. |
Rupees 10,000 is invested at 6% interest compounded annually. How long will it take to accumulate Rs. 20, 000 in the account? |
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Answer» Correct Answer - 12 years Principal amount = Rs. 10,000 Rate of interest = 6% So, amount after one year = Rs. `1.06 xx 1000` Amount after two years = Rs. ` 1.06 xx 1.06 xx 10000` Rs.` (1.06)^(2) xx 10000` and so on So, amount accumulated after n years = Rs. `(1.06)^(n) xx 10000`. Now,` (1.06)^(n) xx 10000 = 20000` ` :. (1.06)^(n) = 2` ` rArr n log_(10)1.06 = log_(10) 2` ` rArr n = (log_(10)2)/(log_(10)1.06) = (0.30103)/(0.025306)` ` rArr log n = log 0.30103 - log 0.025306` ` =-0.5213 - (-1.5968)` ` = 1.0755` ` =11.89` So, it will take approximately 12 years. |
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| 20. |
An initial number of bacteria presented in a culture is 10000. This number doubles every 30 minutes. How long will it take to bacteria to reach the number 100000 ? |
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Answer» Correct Answer - 100 minutes Initially bacteria count is 10000. After 30 min. it doubles i.e., ` 2xx 10000`. After 60 min. it is`2^(2) xx 10000`. So, after t minutes, bacteria count will be ` N = 10000 xx 2^(t/30)` For ` N = 100000`,we have `100000 = 10000 xx 2^(t/30)` ` :. 2^(t/30) = 10` ` t/30 log_(10) 2 = 1` ` rArr t = (30)/(log_(10) 2) = 30/(0.301) = 99. 67` So, it will take approximately 100 minutes. |
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| 21. |
`log_(4) 18 ` isA. a rational numberB. an irrational numberC. a prime numberD. none of these |
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Answer» Correct Answer - B Let ` log_(4) 18 = p//q," where "p,q in I` ` rArr log_(4) 9+log_(4) 2 = p/q` ` rArr 1/2 xx 2 log _(2) 3 + 1/2 = p/q` `rArr log_(2) 3 = p/q - 1/2 = m/n(say)` where m, n` in I and n ne 0` ` rArr 3 = (2) ^(m//n)` ` or 3^(n) = 2^(m) ` (possible when m=n = 0 which is not true) Hence, ` log_(4) 18` is an irrational number. |
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| 22. |
Charles Richter defined the magnitude of an earthquake to be ` M = log_(10) I/S`, where I is the intensity of the earthquake (measured by the amplitude of a seismograph reading taken 100 km from the epicentre of the earthquake) and S is the intensity of a 'standed earthquake' (whose amplitude is 1 micron `=10^(-1)` cm). Each number increase on the Richter scale indicates an intensity ten times stronger. For example. an earthquake of magnitude 5. An earthquake of magnitude 7 is 100 times stronger then an earthquake of magnitude 5. An earthquake of magnitude 8 is 1000 times stronger than an earthquake of magnitude 5. The earthquake in city A registered `8.3` on the Richter scale. In the same year, another earthquake was recorded in city B that was four times stronger. What was the magnitude of the earthquake in city B ? |
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Answer» Correct Answer - ` 8.9020` `M _(A) = log_(10). (I_(A))/S` `:. 8.3 = log_(10). (I_(A))/S` Now `M_(B) = log_(10). (I_(B))/S` Where ` I_(B) = 4I_(A)*` ` :. M_(B) = log_(10). (4I_(A))/S` ` = log_(10) 4+ log_(10). (I_(A))/S` ` = 0.6020 + 8.3` ` = 8.9020` So, magnitude of earthquake in city B is ` 8.9020`. |
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| 23. |
Solve `(log)_(0. 04)(x-1)geq(log)_(0. 2)(x-1)` |
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Answer» ` log_(0.04) (x-1) ge log_(0.2)(x-1)` ` or log_((0.2^(2)))(x-1) ge log_(0.2)(x-1)` ` or 1/2 log_(0.2)(x-1) ge log_(0.2) (x-1)` ` or log_(0.2)(x-1) ge 2 log_(0.2)(x-1)` ` or log_(0.2)(x-1) ge log_(0.2)(x-1)^(2)` ` or (x-1) le (x-1)^(2)` ` or (x-1)^(2)-(x-1) ge 0` ` or (x-1)(x-1-1) ge 0` `or (x-1)(x-2) ge 0` ` or x le 1 or x ge 2` `"Also," x gt 1,` `" Hence, "x ge 2.` |
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| 24. |
Find the value of ` log_(2) (1/(7^(log_(7) 0.125)))`. |
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Answer» Correct Answer - 3 ` log_(2)(1/(7^(log_(2)(2sqrt3)))+2/(log_(3)(2sqrt3)))^(2)=(4log_((2sqrt3))2+ 2log_((2sqrt3))3)^(2)` `= (log_(2sqrt3)16+log_(2sqrt3)9)^(2)` ` = (log_(2sqrt3)144)^(2)` ` (log_(2sqrt3)(2sqrt3)^(4))^(2)` ` = 4^(2) = 16` |
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| 25. |
Find the value of `log_(2sqrt3) 1728`. |
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Answer» Let ` log_((2)sqrt3) 1728 = x` `rArr1728 = (2sqrt3)^(x)` ` rArr(2^(6)3^(3))=(2sqrt3)^(x)` ` rArr (2 sqrt3)^(6) = (2sqrt3)^(x)` ` rArr x = 6` |
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| 26. |
Prove that `2/5 lt log_(10) 3 lt 1/2`. |
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Answer» Let ` log_(10) 3 gt 2/5` ` rArr 3 gt 10^(2/5)` ` rArr 3^(5) gt 10^(2)`, which is true Now, ` log_(10) 3 lt 1/2` ` rArr 3 lt 10^(1/2)` ` rArr 3^(2) lt 10`, which is true Hence, `2/5 lt log_(10) 3 lt 1/2` |
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| 27. |
Show that the number `log_2(7)` is an irrational number |
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Answer» Let us assume given expression is a rational number. Let `log_2(7) = p/q` Here, `p/q` is a rational number. `log_2(7) =p/q=>2^(p/q) = 7` Raising both sides with power of `q`, `2^(p/q**q) = 7^q=>2^p = 7^q` As, `2^p` is always even and `7^q` is always odd. So, our assumption is wrong that given expression is equal to a rational number. Thus, `log_2(7)` is an irrational number. |
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| 28. |
Arrange ` log_(2) 5, log_(0.5) 5, log_(7) 5, log_(3) 5` in decreasing order. |
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Answer» ` log_(2) 5 ` = exponent of 2 for which we get 5 ` log_(7) 5 ` = exponent of 7 for which we get 5 Clearly, ` log_(2) 5 gt log_(7) 5` With similar reasons, we have ` log_(2) 5 gt log_(3) 5 gt log_(7) 5`. Also ` log_(0.5) 5 lt 0` `:. log_(2) 5 gt log_(3) 5 gt log_(7) 5 gt log_(0.5) 5` |
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| 29. |
Which of the following numbers are positive/negative : `(i) log_(sqrt(3))sqrt(2)` `(ii)log_3(4)` |
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Answer» (i) Let ` log_(2)7=xrArr7=2^(x) rArrx gt 0` (ii)Let ` log_(0.2)3= x rArr 3 = 0.2^(x) rArr x lt 0` (iii) Let ` log_(1//3)(1//5)=x rArr 1//5 = (1//3)^(x) rArr 5 = 3^(x) rArr x gt 0` (iv) Let ` log_(4) 3 = x rArr 3 = 4^(x) rArr x lt 0` (v) Let ` log_(2) (log_(2)9) = x rArr log_(2) 9 = 2^(x) rArr 9 =2^(2^(x)) rArr x gt 0` |
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| 30. |
Prove that number ` log_(2)` 7 is an irrational number. |
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Answer» Let ` log_(2) 7 ` is a rational number. Then, ` log_(2)7=p/q , p,q in Q or 7 = 2^(p//q) or 7^(q) = 2^(p)` which is not possible for any integral values of p and q. Hence, ` log_(2) 7 ` is not rational. |
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| 31. |
If ` log_(3) y = x and log_(2) z = x , " find " 72^(x)` in terms of y and z. |
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Answer» ` log_(3) y = x` ` rArr y = 3^(x)` ` log_(2) z=x` ` rArr z = 2^(x)` Now, ` 72^(x) = (2^(3)3^(2))^(x) = 2^(3x)3^(2x) =( 2^(x))^(3)(3^(x))^(2)=y^(3)z^(2)` |
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| 32. |
If sum ` log_(2)x+log_(4) x + log_(16) x + log_(256) x + …=6,` then find the value of x. |
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Answer» `log_(2)x+log_(4)x+log_(16)x+log_(32)x+ * * * = 6` ` rArr log_(2) x+ log_(2^(2))x+log_(2^(4))x+log_(2^(5))x+ * * * = 6` `rArr (1+1/2+1/4+1/8+...)log_(2) x= 6` ` rArr 1/(1-1/2)log_(2) x = 6` ` rArr log_(2) x = 3` ` rArr x = 8` |
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| 33. |
Evaluate `root(7)(0.00003587)`. |
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Answer» Let `x = root(7)(0.00003587).` Then, log `x = 1/7` log (0.00003587) `= 1/7 (bar(5).5548) = 1/7 (-5 + 0.5548)` `= 1/7 (-4.4452) = -0.6350` ` = -1 + (1 - 0.6350) = bar(1).3650` `rArr` x = antilog `(bar(1).3650) = 0.2317.` Hence, the required value is 0.2317. |
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| 34. |
Evaluate `root(3)(0.08034).` |
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Answer» Let `x = root(3)(0.08034).` Then, log `x = 1/3` log (0.08034). `= 1/3 (bar(2).9049) = 1/3(-2 + 0.9049) = 1/3(-1.0951)` `= -0.3650 = -1 + (1 - 0.3650) = bar(1).6350` `rArr` x = antilog `(bar(1).6350) = 0.4315.` |
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| 35. |
Evaluate `(563.4 xx root(3)(0.4773))/((6.15)^(3)).` |
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Answer» Let `x = (563.4 xx root(3)(0.4773))/((6.15)^(3)).` Then, log x = log `563.4 + 1/3` log (0.4573) - 3 log (6.15) `= 2.7508 + 1/3 (bar(1).6602) - 3 xx 0.7889` `= 2.7508 + 1/3 (-3398) - 2.3667` `= 2.7508 - 0.1133 - 2.3667` = 0.2708 `rArr` x = antilog (0.2708) = 1.865. Hence, the value of the given expression is 1.865. |
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| 36. |
Evaluate `sqrt(((76.24)^(5) xx root(3)(65))/((3.2)^(7) xx sqrt(17)))`. |
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Answer» Let `x = sqrt(((76.24)^(5) xx root(3)(65))/((3.2)^(7) xx sqrt(17)))`. Then, log `x = 1/2 * log {((76.24)^(5) xx (65)^(1/3))/((3.2)^(7) xx (17)^(1/2))}` `= 1/2 * {5 log (76.24) + 1/3 log 65 - 7log (3.2) - 1/2 log 17}` `= 1/2 * {5 xx 1.8822 + 1/3 xx 1.8129 - 7 xx 0.5051 - 1/2 xx 1.2304}` `= 1/2 * {9.4110 + 0.6043 - 3.5357 - 0.6152}` `= 1/2 xx (10.0153 - 4.1509) = 1/2 xx 5.8644 = 2.9322` `rArr` x = antilog (2.9322) = 855.5. Hence, the required value of the given expression is 855.5. |
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| 37. |
If ` f.(x)= log((1+x)/(1-x))`, thenA. `f(x_(1))*f(x_(2))=f(x_(1)+x_(2))`B. `f(x+2)-2f(x+1)+f(x)=0`C. `f(x)+f(x+1)=f(x^(2)+x)`D. `f(x_(1))+f(x_(2)) = f ((x_(1)+x_(2))/(1+x_(1)x_(2)))` |
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Answer» Correct Answer - D ` f(x_(1)) + f(x_(2)) = log ((1+x_(1))/(1- x_(1))*(1+x_(2))/(1-x_(2)))` `= log((1+x_(1)x_(2)+x_(1)+x_(2))/(1+x_(1)x_(2)-x_(1)-x_(2)))` `= log((1+(x_(1)+x_(2))/(1+x_(1)x_(2)))/(1-(x_(1)+x_(2))/(1+x_(1)x_(2))))= f ((x_(1)+x_(2))/(1+x_(1)x_(2)))` |
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| 38. |
Solve:`4^(log_2 x)-2x-3=0` |
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Answer» `4^(log_2x) - 2x- 3 = 0` `=>2^(2(log_2x)) - 2x - 3 = 0` `=>2^(log_2x^2) -2x -3 = 0` `=>x^2-2x-3 = 0` (As `a^(log_ab) = b`) `=>x^2 - 3x+x-3 = 0` `=>x(x-3)+1(x-3) = 0` `=>(x-3)(x+1) = 0` `x=3 and x =-1` But, `x` can not be negative, So, `x = 3` |
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| 39. |
Solve `log_2|x-1| |
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Answer» `log_2 |x-1| lt 1` `=>|x-1| lt 2^1` `=>|x-1| lt 2` Case 1: When `x ge1`, Then, `x -1 lt 2` `=> x lt 3` `:. x in [1,3)` Case 2: When `x le 1`, Then, `-x +1 lt 2` `=> -x lt 1` `=> x gt -1` `:. x in (-1,1]` So, common solution for these two cases will be, `x in (-1,3).` |
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| 40. |
The value of ` (sqrt(3+2sqrt2)+sqrt(3-2sqrt2))^(2^(9))` is ________. |
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Answer» Correct Answer - 6 `3+2sqrt2=(sqrt2+1)^(2) and 3-2sqrt2=(sqrt2-1)^(2)` ` rArr log_((sqrt(3+2sqrt2)+sqrt(3-2sqrt2)))2^(9)` ` = 1/(log_(2^(9))((sqrt2+1)+(sqrt2-1)))` ` = 1/(log_(2^(9))2^(3//2))` ` = 9/(3//2)=6` |
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| 41. |
Find the value of ` log_(2) (2root(3)9-2) + log_(2)(12root(3)3+4+4root(3)9)`. |
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Answer» `log_(2)(2root(3)9-2)+log_(2)(12root(3)3+4+4root(3)9)` `=log_(2)((72)^(1//3)-(8)^(1//3)))+log_(2)((72)^(2//3)+(8)^(2//3)+(72)^(1//3)(8)^(1//3))` `=log_(2)((72)^(1//3)-(8)^(1//3)))((72)^(2//3)+(8)^(2//3)+(72)^(1//3)(8)^(1//3))` `= log_(2)(72-8)` ` = log_(2)64=6` |
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| 42. |
What is logarithm of ` 32 root(5) 4" to the base "2sqrt2`? |
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Answer» `log_(2sqrt2)32root(5)4=log_((2^(3//2)))(2^(5)4^(1/5))` ` = log_((2^(3//2)))(2^(5+2/5))` `=2/3 27/5 log_(2) 2` ` 18/5` ` = 3.6` |
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| 43. |
If ` log_(1//2)(4-x)gelog_(1//2)2-log_(1//2)(x-1)`,then x belongs toA. `(1, 2]`B. `[3, 4)`C. `(1, 3]`D. `[1, 4)` |
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Answer» Correct Answer - A::B `log_(1//2)(4-x) ge log_(1//2)2-log_(1//2) (x-1)` `or log_(1//2)(4-x)(x-1) ge log_(1//2) 2` ` or (4-x)(x-1) le 2` ` or x^(2) - 5x + 6 ge 0` ` or (x-3)(x-2) ge 0` ` or x ge 3 or x le 2` But ` x in (1, 4)` ` rArr x in (1, 2] cup [3, 4)` |
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| 44. |
If the equation `x^(log_(a)x^(2))=(x^(k-2))/a^(k),a ne 0`has exactly one solution for x, then the value of k is/areA. `6+4sqrt2`B. `2+6sqrt3)`C. `6-4sqrt2`D. `2-6sqrt3` |
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Answer» Correct Answer - A::C ` (log_(a)x^(2)) log_(a)x =(k-2)log_(a) x - k` (taking log on base a ) Let ` log_(a) x = t`, we get ` 2t^(2) - (k-2) t+k = 0` Putting D= 0 (has only one solution), we have ` (k-2)^(2) - 8k=0` ` or k^(2) - 12k+4 = 0` ` or k=(12 pm sqrt(128))/2` ` or k = 6 pm 4 sqrt2` |
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| 45. |
The value of ` (6 a^(log_(e)b)(log_(a^(2))b)(log_(b^(2))a))/(e^(log_(e)a*log_(e)b))` isA. independent of aB. independent of bC. dependent on aD. dependent on b |
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Answer» Correct Answer - A::B `(6a^(log_(e)b)log_(a^(2))b*log_(b^(2))a)/(e^(log_(e)a*log_(e)b))=(6a^(log_(e)b)1/2log_(a)b*1/2log_(b)a)/((e^(log_(e)a))^(log_(e)b))` ` (6/4 a^(log_(e)b))/(a^(log_(e)b))` `= 3/2` |
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| 46. |
if `log_10 5=a` and `log_10 3=b` then:A. ` log_(30) 8 = (3(1-a))/(b+1) `B. ` log_(40)15= (a+b)/(3-2a)`C. ` log_(243) 32 = (1-a)/b`D. none of these |
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Answer» Correct Answer - A::B::C (1) ` log_(30) 8 = (3log_(10)2)/(log_(10)5+log_(10) 3+log_(10)2)` ` = (3(1-log_(10)5))/(log_(10)5+log_(10)3+log_(10)2)` ` = (3(1-log_(10)5))/(1+log_(10)3) = (3(1-a))/(1+b)` (2)` log_(40) 15=(log_(10)15)/(log_(10)40)` ` = (log_(10)3+log_(10)5)/(log_(10)5+3[1-log_(10)5])=(a+b)/(3-2a)` (3) `log_(243)32 = (log_(10)2)/(log_(10)3)=(1-log_(10)5)/(log_(10)3) = (1-a)/b ` |
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| 47. |
In which of the following, ` m gt n (m,n in R)`?A. `m = (log_(2)5)^(2) and n = log_(2) 20`B. `m = log_(10)2 and n = log_(10) root(3) 10`C. ` m = log_(10) 5 * log_(10) 20 and n = 1`D. ` m = log_(1//2) (1/3) and n = log_(1//3) (1/2)` |
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Answer» Correct Answer - A::D (1) `m - n = (log_(2)5)^(2) - (log_(2)5+2)` ` = (log_(2)5-2)(log_(2)5+1) gt 0` ` :. m gt n ` (2) ` 2^(3) lt 10` ` or 2 lt 10^(1//3)` ` or log_(10) 2 lt log_(10)10^(1//3)` ` :. M lt n` (3) ` m = log_(10)5* log_(10) 20` ` = (log_(10)10-log_(10)2)(log_(10)10+log_(10) 2)` ` (1-log_(10)2)(1+log_(10)2)` ` = 1 - (log_(10)2)^(2)` ` lt 1` ` :. m lt n` (4) ` m = log_(1//2)(1/3) = log_(2) 3 and n = log_(1//3)(1/2) = log_(3) 2` `:. m/n = (log_(2)3)/(log_(3)2) = (log_(2)3)^(2) gt 1` ` :. m gt n ("as " n gt 0)` |
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| 48. |
The set of real values of `x` satisfying the equation`|x-1|^(log_3(x^2)-2log_x(9))=(x-1)^7`A. `1/sqrt3`B. 1C. 2D. 81 |
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Answer» Correct Answer - C::D `|x-1|^(log_(3)x^(2)-2log_(x)9*)= (x-1)^(7)` Since L.H.S. ` gt 0." So, "x gt 1` ` :. (x-1)^(log_(3)x^(2)-2 log_(x)9)=(x-1)^(7)` `rArr x - 1 = 1 or log_(3)x^(2) - 2log_(x)9=7` ` rArr x = 2 or 2 log_(3) x - 4 1/(log_(3)x) - 7 = 0` ` rArr x = 2 or 2(log_(3)x)^(2) - 7 log_(3)x-4 = 0` ` rArr x = 2 or log _(3) x =- 1//2, 4` ` rArr x = 2 or x = 3^(-1//2) , 3^(4)` ` rArr x = 2, 81` |
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| 49. |
If ` a = log_(245) 175 and b = log_(1715) 875,` then the value of `(1-ab)/(a-b)` is ________. |
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Answer» Correct Answer - 5 ` a=(log_(5)175)/(log_(5)245) = (2+log_(5)7)/(1+2 log_(5)7)` ` or a+ 2a log_(5) 7 = 2 + log_(5) 7` ` or log_(5) 7 = (a-2)/(1-2a) ` ...(i) Now ` b = (log_(5)875)/(log_(5)1715) = (3+log_(5)7)/(1+3log_(5)7)` ` or b+3b log_(5) 7 = 3 + log_(5) 7` ` or log_(5) 7 = (b-3)/(1-3b) ` ...(ii) From Eqs. (i) and (ii), we get `(a-2)/(1-2a) = (b-3)/(1-3b)` ` rArr (1-ab)/(a-b) = 5` |
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| 50. |
Solve : `(log)_(0. 3)(x^2-x+1)>0` |
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Answer» `log_(0.3) (x^2-x+1) gt 0` As `0.3 lt 1`, so, sign will change. `=> x^2-x+1 lt (0.3)^0` `=> x^2-x+1 lt 1` `=>x^2-x lt 0` `=>x(x-1) lt 0` `:. x in (0,1)` is the solution for the given equation. |
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