The value of {x in R mid log_{10} ((1.6)^{1-x | vedclass

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(D) For the expression $\log_{10} ((1.6)^{1-x^2} - (0.625)^{6(1+x)})$ to be defined in $R$,the argument of the logarithm must be strictly greater than

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$(-1, 7)$

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(D) For the expression $\log_{10} ((1.6)^{1-x^2} - (0.625)^{6(1+x)})$ to be defined in $R$,the argument of the logarithm must be strictly greater than $0$. $(1.6)^{1-x^2} - (0.625)^{6(1+x)} > 0$ $(1.6)^{1-x^2} > (0.625)^{6(1+x)}$ Since $1.6 = \frac{8}{5}$ and $0.625 = \frac{5}{8} = (\frac{8}{5})^{-1}$,we have: $(\frac{8}{5})^{1-x^2} > ((\frac{8}{5})^{-1})^{6(1+x)}$ $(\frac{8}{5})^{1-x^2} > (\frac{8}{5})^{-6(1+x)}$ Since the base $\frac{8}{5} > 1$,the inequality direction remains the same: $1 - x^2 > -6(1 + x)$ $1 - x^2 > -6 - 6x$ $x^2 - 6x - 7 $(x - 7)(x + 1) Thus,$x \in (-1, 7)$.

The value of $\{x \in R \mid \log_{10} ((1.6)^{1-x^2} - (0.625)^{6(1+x)}) \in R\}$ is

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