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Given that log5 21 =m and log9 75= n show that log5 7 = 1÷2n-1(2mn-m-2)​

User Cacert
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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.

User Teila
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