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Problem What is the product of all solutions to the equation log_(7x)2023*log_(289 x)2023=log_(2023 x)2023 (A) (log_(2023)7*log_(2023)289)^(2) (B) log_(2023)7*log_(2023)289 (C) 1 (D) log_(7)2023*log_(289)2023 (E) (log_(7)2023*log_(289)2023)^(2)

Problem\newlineWhat is the product of all solutions to the equation\newlinelog⁡7x2023⋅log⁡289x2023=log⁡2023x2023 \log _{7 x} 2023 \cdot \log _{289 x} 2023=\log _{2023 x} 2023 \newline(A) (log⁡20237⋅log⁡2023289)2 \left(\log _{2023} 7 \cdot \log _{2023} 289\right)^{2} \newline(B) log⁡20237⋅log⁡2023289 \log _{2023} 7 \cdot \log _{2023} 289 \newline(C) 11\newline(D) log⁡72023⋅log⁡2892023 \log _{7} 2023 \cdot \log _{289} 2023 \newline(E) (log⁡72023⋅log⁡2892023)2 \left(\log _{7} 2023 \cdot \log _{289} 2023\right)^{2}

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Q. Problem\newlineWhat is the product of all solutions to the equation\newlinelog⁡7x2023⋅log⁡289x2023=log⁡2023x2023 \log _{7 x} 2023 \cdot \log _{289 x} 2023=\log _{2023 x} 2023 \newline(A) (log⁡20237⋅log⁡2023289)2 \left(\log _{2023} 7 \cdot \log _{2023} 289\right)^{2} \newline(B) log⁡20237⋅log⁡2023289 \log _{2023} 7 \cdot \log _{2023} 289 \newline(C) 11\newline(D) log⁡72023⋅log⁡2892023 \log _{7} 2023 \cdot \log _{289} 2023 \newline(E) (log⁡72023⋅log⁡2892023)2 \left(\log _{7} 2023 \cdot \log _{289} 2023\right)^{2}
  1. Understand given logarithmic equation: Understand the given logarithmic equation.\newlineThe given equation is log⁡7x2023×log⁡289x2023=log⁡2023x2023\log_{7x}2023 \times \log_{289x}2023 = \log_{2023x}2023. We need to find the product of all solutions to this equation.
  2. Use change of base formula: Use the change of base formula to rewrite the logarithms in a common base.\newlineChange of Base Formula: log⁡b(a)=log⁡c(a)log⁡c(b)\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\newlineWe can rewrite each term using the change of base formula with a common base, for example, base 20232023.\newlinelog⁡(7x)(2023)=log⁡(2023)log⁡(7x)\log_{(7x)}(2023) = \frac{\log(2023)}{\log(7x)}\newlinelog⁡(289x)(2023)=log⁡(2023)log⁡(289x)\log_{(289x)}(2023) = \frac{\log(2023)}{\log(289x)}\newlinelog⁡(2023x)(2023)=log⁡(2023)log⁡(2023x)\log_{(2023x)}(2023) = \frac{\log(2023)}{\log(2023x)}
  3. Substitute rewritten logarithms: Substitute the rewritten logarithms back into the equation.\newline(log⁡(2023)/log⁡(7x))⋅(log⁡(2023)/log⁡(289x))=log⁡(2023)/log⁡(2023x)(\log(2023) / \log(7x)) \cdot (\log(2023) / \log(289x)) = \log(2023) / \log(2023x)
  4. Cross-multiply to simplify: Cross-multiply to simplify the equation. log⁡(2023)2=log⁡(7x)⋅log⁡(289x)⋅log⁡(2023x)\log(2023)^2 = \log(7x) \cdot \log(289x) \cdot \log(2023x)
  5. Recognize form of product of logarithms: Recognize that the equation has the form of a product of logarithms equal to a constant. This suggests that the solutions for xx will be such that 7x7x, 289x289x, and 2023x2023x are powers of 20232023, since the base of the logarithm on the right side of the equation is 20232023.
  6. Solve for x by equating: Solve for xx by equating each term to 20232023.\newlineSince 7x7x, 289x289x, and 2023x2023x must be powers of 20232023, we can write:\newline7x=2023a7x = 2023^a\newline289x=2023b289x = 2023^b\newline2023x=2023c2023x = 2023^c\newlineWhere aa, 2023202300, and 2023202311 are integers.
  7. Recognize 2023x=202312023x = 2023^1: Recognize that 2023x=202312023x = 2023^1, which implies x=1x = 1. From the third equation, we immediately see that xx must be 11.
  8. Substitute x=1x = 1 for consistency: Substitute x=1x = 1 back into the other equations to check for consistency.\newline7x=7(1)=7=2023a7x = 7(1) = 7 = 2023^a\newline289x=289(1)=289=2023b289x = 289(1) = 289 = 2023^b\newlineWe see that aa and bb must be such that 77 and 289289 are powers of 20232023, which is not possible since 20232023 is not a power of 77 or 289289. Therefore, the only solution for x=1x = 122 is x=1x = 133.
  9. Determine product of all solutions: Determine the product of all solutions.\newlineSince the only solution is x=1x = 1, the product of all solutions is simply 11.

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