For the inequality 2^{log_{sqrt{2}}(x - 1)} | vedclass

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(B) Given the inequality $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$. First,consider the domain of the logarithmic function: $x - 1 > 0 \implies x > 1$. Usin

Vedclass · Accepted Answer

$(4, +\infty)$

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(B) Given the inequality $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$. First,consider the domain of the logarithmic function: $x - 1 > 0 \implies x > 1$. Using the property $\log_{a^n}(b) = \frac{1}{n} \log_a(b)$,we have $\log_{\sqrt{2}}(x - 1) = \log_{2^{1/2}}(x - 1) = 2 \log_2(x - 1) = \log_2((x - 1)^2)$. Substituting this into the inequality: $2^{\log_2((x - 1)^2)} > x + 5$. Since $2^{\log_2(y)} = y$,the inequality becomes $(x - 1)^2 > x + 5$. Expanding the left side: $x^2 - 2x + 1 > x + 5$. Rearranging the terms: $x^2 - 3x - 4 > 0$. Factoring the quadratic: $(x - 4)(x + 1) > 0$. The roots are $x = 4$ and $x = -1$. The inequality holds for $x \in (-\infty, -1) \cup (4, +\infty)$. Considering the domain $x > 1$,the intersection of $(x > 1)$ and $(x  4)$ is $x > 4$. Thus,the solution set is $(4, +\infty)$.

For the inequality $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$,the set of real values of $x$ is:

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