Asked by Anonymous

Express log A in terms of the
2
logarithms of prime numbers:


A= (sqrt(3)*sqrt4(125))/7^3

Answers

Answered by Steve
Assuming that sqrt4(125) means ∜125, then

A = 3^(1/2) * 5^(3/4) / 7^3
so,
logA = (1/2)log3 + (3/4)log5 - 3log7

The base of the logs does not matter. The rules still hold.
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