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

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Question

(A) For the logarithm $\log_{10}(A)$ to be defined in $R$, the argument $A$ must be strictly greater than $0$.Thus, $(1.6)^{1-x^2} - (0.625)^{6(1+x)}

Vedclass · Accepted Answer

$(-1, 7)$

Vedclass · Answer

(A) For the logarithm $\log_{10}(A)$ to be defined in $R$, the argument $A$ must be strictly greater than $0$.Thus, $(1.6)^{1-x^2} - (0.625)^{6(1+x)} > 0$.Note that $1.6 = \frac{8}{5}$ and $0.625 = \frac{5}{8} = (\frac{8}{5})^{-1}$.So, $(\frac{8}{5})^{1-x^2} > (\frac{8}{5})^{-6(1+x)}$.Since the base $\frac{8}{5} > 1$, the inequality holds for the exponents:$1 - x^2 > -6(1 + x)$$1 - x^2 > -6 - 6x$$x^2 - 6x - 7 Factoring the quadratic: $(x - 7)(x + 1) The solution to this inequality is $x \in (-1, 7)$.

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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