If $l o g_{10} 5 + l o g_{10} (5 x + 1) = l o g_{10} (x + 5) + 1, t h e n (x^{2} - 9)$ is equal to:

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log₁₀5

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Solution

The correct option is D
0

$l o g_{10} 5 + l o g_{10} (5 x + 1) = l o g_{10} (x + 5) + 1 (G i v e n) \Rightarrow l o g_{10} 5 + l o g_{10} (5 x + 1) = l o g_{10} (x + 5) + l o g_{10} 10 (l o g_{10} 10 = 1) \Rightarrow l o g_{10} [5 (5 x + 1)] = l o g_{10} [10 (x + 5)] (l o g A + l o g B = l o g (A B)) \Rightarrow 5 (5 x + 1) = 10 (x + 5) \Rightarrow 5 x + 1 = 2 x + 10 d i v i d i n g b y 5 o n b o t h t h e s i d e s \Rightarrow 3 x = 9 \Rightarrow x = 3 \Rightarrow (x^{2} - 9) = 9 - 9 = 0$

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