The product of all positive real values of x | vedclass

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(B) Given equation: $x^{16(\log_5 x)^3 - 68 \log_5 x} = 5^{-16}$.Taking $\log_5$ on both sides:$(16(\log_5 x)^3 - 68 \log_5 x) \cdot \log_5 x = \log_5

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$1$

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(B) Given equation: $x^{16(\log_5 x)^3 - 68 \log_5 x} = 5^{-16}$.Taking $\log_5$ on both sides:$(16(\log_5 x)^3 - 68 \log_5 x) \cdot \log_5 x = \log_5(5^{-16})$.Let $t = \log_5 x$. Then the equation becomes:$(16t^3 - 68t) \cdot t = -16$.$16t^4 - 68t^2 + 16 = 0$.Dividing by $4$:$4t^4 - 17t^2 + 4 = 0$.Let $u = t^2$. Then $4u^2 - 17u + 4 = 0$.$(4u - 1)(u - 4) = 0$.So,$u = 1/4$ or $u = 4$.Since $u = t^2 = (\log_5 x)^2$,we have $(\log_5 x)^2 = 1/4$ or $(\log_5 x)^2 = 4$.This gives $\log_5 x = \pm 1/2$ or $\log_5 x = \pm 2$.The values of $x$ are $5^{1/2}, 5^{-1/2}, 5^2, 5^{-2}$.The product of these values is $5^{1/2 - 1/2 + 2 - 2} = 5^0 = 1$.

The product of all positive real values of $x$ satisfying the equation $x^{(16(\log_5 x)^3 - 68 \log_5 x)} = 5^{-16}$ is:

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