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So. option (b) is incorrect.
If log 2 = 0.3010 and log 3 = 0.4771, the value of log_{5} 512 is:
Given, log 27 = 1.431
⇒ log (3^{3}) = 1.431
⇒ 3 log 3 = 1.431
⇒ log 3 = 0.477
∴ log 9 = log (3^{2}) = 2 log 3
⇒ (2 x 0.477) = 0.954
⇒  log_{10} (7 x 10)
⇒  (log_{10} 7 + log_{10} 10)
⇒  (a + 1)
log_{10} 80 = log_{10} (8 x 10)
⇒ log_{10} 8 + log_{10} 10
⇒ log_{10}(2^{3}) + 1
⇒ 3 log_{10} 2 + 1
⇒ (3 x 0.3010) + 1
⇒ 1.9030
If log_{10} 5 + log_{10} (5x + 1) = log_{10} (x + 5) + 1, then x is equal to:
log_{10} 5 + log_{10} (5x + 1) = log_{10} (x + 5) + 1
⇒ log_{10} 5 + log_{10} (5x + 1) = log_{10} (x + 5) + log_{10} 10
⇒ log_{10} [5 (5x + 1)] = log_{10} [10(x + 5)]
⇒ 5(5x + 1) = 10(x + 5)
⇒ 5x + 1 = 2x + 10
⇒ 3x = 9
⇒ x = 3.
Given expression = 1/log_{60} 3 + 1/log_{60} 4 + 1/log_{60} 5
= log_{60} (3 x 4 x 5)
= log_{60} 60
= 1.
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