show that log 10800 = 4 log 2 + 3 log 3 + 2 log 5
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lhs = log 10800
= log(2*2*2*2*3*3*3*5*5) [ write 10800 product of prime factors]
= log (2^4)* (3^3)*5^2
= log 2^4 + log 3^3 + log 5^2 [ ∵ log xyz = log x +log y +log z]
= 4log2 + 3 log 3 + 2 log 5 [ ∵ log a^n = n log a ]
= RHS
= log(2*2*2*2*3*3*3*5*5) [ write 10800 product of prime factors]
= log (2^4)* (3^3)*5^2
= log 2^4 + log 3^3 + log 5^2 [ ∵ log xyz = log x +log y +log z]
= 4log2 + 3 log 3 + 2 log 5 [ ∵ log a^n = n log a ]
= RHS
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This is chemistry question.
Que: Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
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