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.
Express log A in terms of the
2
logarithms of prime numbers:
A= (sqrt(3)*sqrt4(125))/7^3
1 answer
1 answer