The domain of the function f(x) = log_{3+x}(x | vedclass

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Question

(C) For the function $f(x) = \log_{3+x}(x^2 - 1)$ to be defined,the following conditions must be satisfied:$1$. The argument of the logarithm must be

Vedclass · Accepted Answer

$(-3, -2) \cup (-2, -1) \cup (1, \infty)$

Vedclass · Answer

(C) For the function $f(x) = \log_{3+x}(x^2 - 1)$ to be defined,the following conditions must be satisfied:$1$. The argument of the logarithm must be positive: $x^2 - 1 > 0 \implies x^2 > 1 \implies x \in (-\infty, -1) \cup (1, \infty)$.$2$. The base of the logarithm must be positive and not equal to $1$: $3 + x > 0 \implies x > -3$ and $3 + x 
eq 1 \implies x 
eq -2$.Combining these conditions:$x > -3$ and $x 
eq -2$ and $(x  1)$.Intersection of these sets:$(-3, -2) \cup (-2, -1) \cup (1, \infty)$.

The domain of the function $f(x) = \log_{3+x}(x^2 - 1)$ is

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