Problem:
What is the product of all solutions to the equation
log7x​2023⋅log289x​2023=log2023x​2023?
Answer Choices:
A. (log2023​7⋅log2023​289)2
B. log2023​7⋅log2023​289
C. 1
D. log7​2023⋅log289​2023
E. (log7​2023⋅log289​2023)2
Solution:
Changing the logarithms to base b=2023=7.289 using the Change of Base Formula gives
logb​7x1​⋅logb​289x1​=logb​bx1​.
It follows that logb​7x⋅logb​289x=logb​bx. Let y=logb​x. It is possible to rewrite this equation as
(y+logb​7)(y+logb​289)=y+1.
Note that logb​7+logb​289=1. Expanding the quadratic equation and simplifying gives
y2+y+(logb​7⋅logb​289)=y+1.
Then y2=1−logb​7⋅logb​289>0, so there are two solutions for y, which can be denoted as y1​ and y2​=−y1​. Setting xi​=byi​ (for i=1,2 ) gives x1​x2​=by1​+y2​=b0=1. Hence the requested product is (C)1​.
Note: The solutions to the given equation are approximately 943.8 and its reciprocal.
The problems on this page are the property of the MAA's American Mathematics Competitions