If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:
[A].
[B].
[C].
[D].
Answer: Option C
Explanation:
| log5 512 |
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| = 3.876 |
[B].
[C].
[D].
Answer: Option C
Explanation:
| log5 512 |
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| = 3.876 |
[B].
[C].
[D].
Answer: Option B
Explanation:
| log10 80 | = log10 (8 x 10) |
| = log10 8 + log10 10 | |
| = log10 (23 ) + 1 | |
| = 3 log10 2 + 1 | |
| = (3 x 0.3010) + 1 | |
| = 1.9030. |
| 699 |
| 301 |
[B].
| 1000 |
| 301 |
[C].
[D].
Answer: Option B
Explanation:
| log2 10 = | 1 | = | 1 | = | 10000 | = | 1000 | . |
| log10 2 | 0.3010 | 3010 | 301 |
[B].
[C].
[D].
Answer: Option C
Explanation:
log 27 = 1.431
log (33 ) = 1.431
3 log 3 = 1.431
log 3 = 0.477
log 9 = log(32 ) = 2 log 3 = (2 x 0.477) = 0.954.
[B].
[C].
[D].
Answer: Option B
Explanation:
log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10
log10 [5 (5x + 1)] = log10 [10(x + 5)]
5(5x + 1) = 10(x + 5)
5x + 1 = 2x + 10
3x = 9
x = 3.
| log | a | = | x |
| b | y |
[B].
| log a | = | x |
| log b | y |
[C].
| log a | = | y |
| log b | x |
[D].
Answer: Option C
Explanation:
ax = by
log ax = log by
x log a = y log b
| log a | = | y | . | |
| log b | x |
[B].
[C].
[D].
Answer: Option B
Explanation:
(a) Since logaa = 1, so log10 10 = 1.
(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3
log (2 + 3) log (2 x 3)
(c) Since loga 1 = 0, so log10 1 = 0.
(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.
So, (b) is incorrect.
[B].
[C].
[D].
Answer: Option C
Explanation:
log 2x = 10 x = 210.
logxy = 100
y = x100
y = (210)100 [put value of x]
y = 21000.
| 1 |
| 8 |
[B].
[C].
[D].
Answer: Option B
Explanation:
Let log2 16 = n.
Then, 2n = 16 = 24 n = 4.
log2 16 = 4.
[B].
[C].
[D].
Answer: Option D
Explanation:
One of AB, AD and CD must have given.
So, the data is inadequate.