Maths Olympiad Prep

Track / Stage 4 / 178 of 340 #438 of 1964

Problem 438

AMC 12 late, AIME early
Algebra Difficulty 4.8 Multiple choice

8248 \cdot 24 satisfies the equation (log3x)(logx5)=log35\left(\log _{3} x\right)\left(\log _{x} 5\right)=\log _{3} 5 for the positive number xx

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Official solution

[Solution] If x>0x>0 and x1x \neq 1, let a=log3x,b=logx5a=\log _{3} x, \quad b=\log _{x} 5 and c=log35c=\log _{3} 5, then x=3a,5=xbx=3^{a}, \quad 5=x^{b} and 5=3c5=3^{c}.

Thus 3ab=3c3^{a b}=3^{c} or ab=ca b=c.

That is, when x1x \neq 1, ab=ca b=c.
Therefore, the answer is (D)(D).

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.