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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 $\newline$ What is the product of all solutions to the equation $\newline$ $\log _{7 x} 2023 \cdot \log _{289 x} 2023=\log _{2023 x} 2023$ $\newline$ (A) $\left(\log _{2023} 7 \cdot \log _{2023} 289\right)^{2}$ $\newline$ (B) $\log _{2023} 7 \cdot \log _{2023} 289$ $\newline$ (C) $1$ $\newline$ (D) $\log _{7} 2023 \cdot \log _{289} 2023$ $\newline$ (E) $\left(\log _{7} 2023 \cdot \log _{289} 2023\right)^{2}$

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Product property of logarithms

Full solution

Q. Problem

\newline

What is the product of all solutions to the equation

\newline

\log _{7 x} 2023 \cdot \log _{289 x} 2023=\log _{2023 x} 2023

\newline

(A)

\left(\log _{2023} 7 \cdot \log _{2023} 289\right)^{2}

\newline

(B)

\log _{2023} 7 \cdot \log _{2023} 289

\newline

(C)

1

\newline

(D)

\log _{7} 2023 \cdot \log _{289} 2023

\newline

(E)

\left(\log _{7} 2023 \cdot \log _{289} 2023\right)^{2}

Understand given logarithmic equation: Understand the given logarithmic equation. $\newline$ The given equation is $\log_{7x}2023 \times \log_{289x}2023 = \log_{2023x}2023$ . We need to find the product of all solutions to this equation.
Use change of base formula: Use the change of base formula to rewrite the logarithms in a common base. $\newline$ Change of Base Formula: $\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$ $\newline$ We can rewrite each term using the change of base formula with a common base, for example, base $2023$ . $\newline$ $\log_{(7x)}(2023) = \frac{\log(2023)}{\log(7x)}$ $\newline$ $\log_{(289x)}(2023) = \frac{\log(2023)}{\log(289x)}$ $\newline$ $\log_{(2023x)}(2023) = \frac{\log(2023)}{\log(2023x)}$
Substitute rewritten logarithms: Substitute the rewritten logarithms back into the equation. $\newline$ $(\log(2023) / \log(7x)) \cdot (\log(2023) / \log(289x)) = \log(2023) / \log(2023x)$
Cross-multiply to simplify: Cross-multiply to simplify the equation. $\log(2023)^2 = \log(7x) \cdot \log(289x) \cdot \log(2023x)$
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 $x$ will be such that $7x$ , $289x$ , and $2023x$ are powers of $2023$ , since the base of the logarithm on the right side of the equation is $2023$ .
Solve for x by equating: Solve for $x$ by equating each term to $2023$ . $\newline$ Since $7x$ , $289x$ , and $2023x$ must be powers of $2023$ , we can write: $\newline$ $7x = 2023^a$ $\newline$ $289x = 2023^b$ $\newline$ $2023x = 2023^c$ $\newline$ Where $a$ , $2023$ $0$ , and $2023$ $1$ are integers.
Recognize $2023x = 2023^1$ : Recognize that $2023x = 2023^1$ , which implies $x = 1$ . From the third equation, we immediately see that $x$ must be $1$ .
Substitute $x = 1$ for consistency: Substitute $x = 1$ back into the other equations to check for consistency. $\newline$ $7x = 7(1) = 7 = 2023^a$ $\newline$ $289x = 289(1) = 289 = 2023^b$ $\newline$ We see that $a$ and $b$ must be such that $7$ and $289$ are powers of $2023$ , which is not possible since $2023$ is not a power of $7$ or $289$ . Therefore, the only solution for $x = 1$ $2$ is $x = 1$ $3$ .
Determine product of all solutions: Determine the product of all solutions. $\newline$ Since the only solution is $x = 1$ , the product of all solutions is simply $1$ .

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\newline

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