Question

What is the product of all the solutions to the equation $\begin{gathered} \log_{7x}2023\cdot\log_{289x}2023\\ {}=\log_{2023x}2023? \end{gathered}$

$(\log_{2023}7\cdot\log_{2023}289)^2$

$\log_{2023}7\cdot\log_{2023}289$

$1$

$\log_7 2023\cdot\log_{289}2023$

$(\log_7 2023\cdot\log_{289}2023)^2$

Answer: C

Concepts:logarithm Vieta’s Formulas

Difficulty rating: 2040

Solution:

Let $a=\log_{2023}7$ and $b=\log_{2023}289.$ Since $2023=7\cdot 289,$ we have $a+b=1.$ Writing $t=\log_{2023}x,$ each logarithm becomes a reciprocal, and the equation turns into $(1+t)=(a+t)(b+t).$

Expanding and using $a+b=1,$ the linear terms cancel, leaving $t^2+(ab-1)=0.$ Its two roots satisfy $t_1+t_2=0.$

The corresponding solutions multiply to $x_1x_2=2023^{t_1}\cdot 2023^{t_2}$ $=2023^{\,t_1+t_2}$ $=2023^0=1.$

Thus, the correct answer is C.

2023 AMC 12A Problem 19

Answer: C

Difficulty rating: 2040

Solution:

Problem 19 in Other Years