Answer: To show that log5 7 = 1/(2n-1)(2mn-m-2), we'll start by using logarithmic properties and the given information.
First, let's express log9 75 in terms of the base 5 logarithm, using the change of base formula:
log9 75 = log5 75 / log5 9
Next, let's simplify the expression inside the logarithm by breaking down 75 and 9 into their prime factors:
log5 (3^2 * 5^2) / log5 (3^2)
Now, using logarithmic properties, we can split the logarithm of a product into the sum of logarithms:
log5 (3^2) + log5 (5^2) - log5 (3^2)
Simplifying further, we have:
2 log5 3 + 2 log5 5 - 2 log5 3
The 2 log5 3 and -2 log5 3 terms cancel each other out, leaving:
2 log5 5
Now, let's substitute the value of m from the given information (log5 21 = m) into the expression:
2 log5 5 = 2 (log5 (5^2)) = 2(2 log5 5) = 4 log5 5 = 4m
Now, let's substitute the value of n from the given information (log9 75 = n) into the expression:
4m = 4(log5 5) = 4(1/2 log5 (5^2)) = 4(1/2 log5 25) = 4(1/2 log5 (5^2 * 5^2))
Using logarithmic properties, we can split the logarithm of a product into the sum of logarithms:
4(1/2 (log5 5^2 + log5 5^2)) = 4(1/2 (2 log5 5 + 2 log5 5)) = 4(1/2 (4 log5 5))
Simplifying further, we have:
4(1/2) (4 log5 5) = 4(2 log5 5) = 8 log5 5 = 8n
Finally, substituting the values of m and n into the expression:
log5 7 = 1/(2n-1)(2mn-m-2) = 1/(2(8) - 1)(2(4m) - m - 2) = 1/(16 - 1)(8m - m - 2) = 1/15(7m - 2)
Therefore, we have shown that log5 7 = 1/(2n-1)(2mn-m-2) = 1/15(7m - 2), using the given values of log5 21 = m and log9 75 = n.