| Q. No. 1: | Find the value of 2log35 - 5log32
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| A : | 1/2
| | B : | 3/2
| | C : | 1
| | D : | 0
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Answer: D
As, alogbc = clogba
Thus, 5log32 = 2log35
The both terms are equal hence 0 is the answer.
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| Q. No. 2: | Find the value of log(32)(54) * log(52)(34)
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| A : | 5
| | B : | 3
| | C : | 4
| | D : | 2
| | |
Answer: C
4/2 *log35 * 4/2 * log53 = 4(log35 * log53) = 4.
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| Q. No. 3: | Find the simplest value of
log7√7√7√7....infinite
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| A : | 0
| | B : | 3
| | C : | 1/2
| | D : | 1
| | |
Answer: D
Let x = √7√7√7√7...infinite
=> x2 = 7x
=> x=7 or x=0.
log7√7√7√7√7...infinite = log7x = log77 = 1.
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| Q. No. 4: | If x= 1+loga bc, y = 1+ logb ca, z= 1+ logc ab, then which of the following option is hold true? |
| A : | xyz = xy+yz+zx
| | B : | xyz = 2x + yz
| | C : | xyz = 2zx/(x+y)
| | D : | xyz = xz/(xy+yz)
| | |
Answer: A
x = 1 + loga bc = loga a + loga bc = loga abc
Similarly, y = logb abc and z = logc abc
Thus, 1/x + 1/y + 1/z = 1.
=> xy + yz +zx = xyz
| |
| Q. No. 5: | Find the value of x in
logx 2 * log(1/16) 2 = log(x/64) 2
|
| A : | 2
| | B : | 1/2
| | C : | 8
| | D : | 7
| | |
Answer: C
1/log2 x * 1/log2 (x/16) = 1/log2 (x/64)
=> log2 x*log2 (x/16) = log2 (x/64)
=> (log2 x)2 - 5log2 x + 6 =0
=> (log2 x - 3)(log2 x - 2) =0
log2 x =3 and log2 x =2
=> x= 8 and x=4 | |
| Q. No. 6: | If (log a)/(b-c) = (log b)/(c-a) = (log c)/(a-b), then which of the following options holds true?
|
| A : | abbcca =1
| | B : | aabbcc =1
| | C : | a2ab2bc2c =1
| | D : | aabbbccac =1
| | |
Answer: B
Let log a/(b-c) = log b/(c-a) = log c/(a-b) =r
aloga + blogb + clogc = r (a(b-c) + b(c-a) + c(a-b)) = r(0) = 0
Thus, log aa + log bb + log cc =0
=> log (aabbcc )= 0 = log 1
aabbcc =1
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