2023 AMC 12A Problem 19

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

Convert to base

2023

and set

t=\log_{2023}x;

note

2023=7\cdot 17^2,

so

\log_{2023}7+\log_{2023}289=1

The equation becomes a quadratic in

t

whose two roots sum to

0

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.

Problem 19 in other years