The set of real values of x for which 2^{log_ | vedclass

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(B) Given the inequality: $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$First,note the domain condition for the logarithm: $x - 1 > 0 \Rightarrow x > 1$.Using t

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$(4, \infty )$

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(B) Given the inequality: $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$First,note the domain condition for the logarithm: $x - 1 > 0 \Rightarrow x > 1$.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 back into the inequality: $2^{\log_2((x - 1)^2)} > x + 5$.Using the identity $a^{\log_a(y)} = y$,we get: $(x - 1)^2 > x + 5$.Expanding the expression: $x^2 - 2x + 1 > x + 5 \Rightarrow x^2 - 3x - 4 > 0$.Factoring the quadratic: $(x - 4)(x + 1) > 0$.The solution to this inequality is $x  4$.Considering the domain condition $x > 1$,the intersection of $x > 1$ and $(x  4)$ is $x > 4$.Thus,the set of real values is $(4, \infty )$.

The set of real values of $x$ for which $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$ is

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