Problem:
What is the product of all solutions to the equation
log7x2023⋅log289x2023=log2023x2023?
Answer Choices:
A. (log20237⋅log2023289)2
B. log20237⋅log2023289
C. 1
D. log72023⋅log2892023
E. (log72023⋅log2892023)2
Solution:
Changing the logarithms to base b=2023=7.289 using the Change of Base Formula gives
logb7x1⋅logb289x1=logbbx1.
It follows that logb7x⋅logb289x=logbbx. Let y=logbx. It is possible to rewrite this equation as
(y+logb7)(y+logb289)=y+1.
Note that logb7+logb289=1. Expanding the quadratic equation and simplifying gives
y2+y+(logb7⋅logb289)=y+1.
Then y2=1−logb7⋅logb289>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 x1x2=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