If log | a | + | log | b | = log (a + b), then: |
b | a |
log | a | + log | b | = log (a + b) |
b | a |
⟹ log (a + b) = log | ❨ | a | x | b | ❩ | = log 1. |
b | a |
So, a + b = 1.
Time taken to travel 68 km downstream = | ❨ | 68 | ❩hrs = 4 hrs. |
17 |
Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.
⟹ | ❨ | x x 8 | 4 | ❩ | = (y x 4) |
5 |
⟹ | 44 | x =4y |
5 |
⟹ y = | 11 | x. |
5 |
∴ Required ratio = | ❨ | y + x | ❩ | : | ❨ | y - x | ❩ |
2 | 2 |
= | ❨ | 16x | x | 1 | ❩ | : | ❨ | 6x | x | 1 | ❩ |
5 | 2 | 5 | 2 |
= | 8 | : | 3 |
5 | 5 |
= 8 : 3.
Then, his rate downstream = 2x kmph.
∴ (Speed in still water) : (Speed of stream) = | ❨ | 2x + x | ❩ | : | ❨ | 2x - x | ❩ |
2 | 2 |
= | 3x | : | x |
2 | 2 |
= 3 : 1.
Speed in still water = | 1 | (11 + 5) kmph = 8 kmph. |
2 |
Speed downstream = 10.5 kmph.
∴ Total time taken = | ❨ | 105 | + | 105 | ❩hours = 24 hours. |
7.5 | 10.5 |
log | a | = | x |
b | y |
log a | = | x |
log b | y |
log a | = | y |
log b | x |
log a | = | y |
log b | x |
⟹ log ax = log by
⟹ x log a = y log b
⟹ | log a | = | y | . |
log b | x |
699 |
301 |
1000 |
301 |
1000 |
301 |
log2 10 = | 1 | = | 1 | = | 10000 | = | 1000 | . |
log10 2 | 0.3010 | 3010 | 301 |
log10 80 | = log10 (8 x 10) |
= log10 8 + log10 10 | |
= log10 (23 ) + 1 | |
= 3 log10 2 + 1 | |
= (3 x 0.3010) + 1 | |
= 1.9030. |
1 |
8 |
Then, 2n = 16 = 24 ⟹ n = 4.
∴ log2 16 = 4.
⟹ 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.
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