Question 1209787: If 5^x = 7^y = 1225
find xy/(x + y)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Given that $5^x = 7^y = 1225$, we can use the following steps to find $\frac{xy}{x+y}$.
1. Express 1225 as a product of its prime factors.
2. Substitute the prime factorization of 1225 into the given equations.
3. Express x and y in terms of the exponents of the prime factors.
4. Substitute the expressions for x and y into the expression $\frac{xy}{x+y}$.
5. Simplify the expression.
Here are the detailed steps:
1. Express 1225 as a product of its prime factors.
$1225 = 5^2 \cdot 7^2$
2. Substitute the prime factorization of 1225 into the given equations.
$5^x = 5^2 \cdot 7^2$
$7^y = 5^2 \cdot 7^2$
3. Express x and y in terms of the exponents of the prime factors.
$x = 2 + 2 \log_5 7$
$y = 2 + 2 \log_7 5$
4. Substitute the expressions for x and y into the expression $\frac{xy}{x+y}$.
$\frac{xy}{x+y} = \frac{(2 + 2 \log_5 7)(2 + 2 \log_7 5)}{(2 + 2 \log_5 7) + (2 + 2 \log_7 5)}$
5. Simplify the expression.
$\frac{xy}{x+y} = \frac{4 + 4 \log_5 7 + 4 \log_7 5 + 4 \log_5 7 \log_7 5}{4 + 2 \log_5 7 + 2 \log_7 5}$
$\frac{xy}{x+y} = \frac{2 + 2 \log_5 7 + 2 \log_7 5 + 2 \log_5 7 \log_7 5}{2 + \log_5 7 + \log_7 5}$
$\frac{xy}{x+y} = \frac{2 (1 + \log_5 7 + \log_7 5 + \log_5 7 \log_7 5)}{2 + \log_5 7 + \log_7 5}$
$\frac{xy}{x+y} = \frac{2 (1 + \log_5 7)(1 + \log_7 5)}{2 + \log_5 7 + \log_7 5}$
Since $\log_a b = \frac{1}{\log_b a}$, we have $\log_5 7 \log_7 5 = 1$. Therefore,
\[\frac{xy}{x + y} = \frac{2 (1 + \log_5 7)(1 + \log_7 5)}{2 + \log_5 7 + \log_7 5} = \frac{2 (1 + \log_5 7 + \log_7 5 + 1)}{2 + \log_5 7 + \log_7 5} = \boxed{2}.\]
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