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Let x and y be positive real numbers such that
$$\log_{5}{(x + y)} + \log_{5}{(x - y)} = 3,$$ and $$\log_{2}{y} - \log_{2}{x} = 1 - \log_{2}{3}$$. Then $$xy$$ equals
We have, $$\log_{5}{(x + y)} + \log_{5}{(x - y)} = 3$$
=> $$x^2-y^2=125$$......(1)
$$\log_{2}{y} - \log_{2}{x} = 1 - \log_{2}{3}$$
=>$$\ \frac{\ y}{x}$$ = $$\ \frac{\ 2}{3}$$
=> 2x=3y => x=$$\ \frac{\ 3y}{2}$$
On substituting the value of x in 1, we get
$$\ \frac{\ 5x^2}{4}$$=125
=>y=10, x=15
Hence xy=150
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