⟹ 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.
1 |
8 |
Then, 2n = 16 = 24 ⟹ n = 4.
∴ log2 16 = 4.
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 |
1000 |
301 |
1000 |
301 |
log2 10 = | 1 | = | 1 | = | 10000 | = | 1000 | . |
log10 2 | 0.3010 | 3010 | 301 |
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 |
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.
(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.
log (264) | = 64 x log 2 |
= (64 x 0.30103) | |
= 19.26592 |
Its characteristic is 19.
Hence, then number of digits in 264 is 20.
log5 512 |
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= 3.876 |
If log10 7 = a, then log10 | ❨ | 1 | ❩ | is equal to: |
70 |
a |
10 |
1 |
10a |
|
= log10 1 - log10 70 | |||||
= - log10 (7 x 10) | ||||||
= - (log10 7 + log10 10) | ||||||
= - (a + 1). |
∴ logx y = 100
⟹ y = x100
⟹ y = (210)100 [put value of x]
⟹ y = 21000.
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