The domain of f(x) = log left[(2.5)^{3-x^2} - | vedclass

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(B) The function $f(x) = \log \left[(2.5)^{3-x^2} - (0.4)^{x+9}\right]$ is defined if the argument of the logarithm is strictly positive: $(2.5)^{3-x^

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$(-3, 4)$

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(B) The function $f(x) = \log \left[(2.5)^{3-x^2} - (0.4)^{x+9}\right]$ is defined if the argument of the logarithm is strictly positive: $(2.5)^{3-x^2} - (0.4)^{x+9} > 0$ $\Rightarrow (2.5)^{3-x^2} > (0.4)^{x+9}$ Since $0.4 = \frac{4}{10} = \frac{2}{5} = (2.5)^{-1}$,we can write: $(2.5)^{3-x^2} > (2.5)^{-(x+9)}$ Since the base $2.5 > 1$,the inequality holds for the exponents: $3 - x^2 > -x - 9$ $x^2 - x - 12 $(x - 4)(x + 3) This inequality holds for $x \in (-3, 4)$.

The domain of $f(x) = \log \left[(2.5)^{3-x^2} - (0.4)^{x+9}\right]$ is

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