1983 AIME Problem 1

Attempt Problem 1 of the 1983 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1983 AIME solutions, or check the answer key.

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1.

Let x,x, y,y, and zz all exceed 11 and let ww be a positive number such that logxw=24,logyw=40,logxyzw=12. \begin{aligned} \log_x w &= 24,\\ \log_y w &= 40,\\ \log_{xyz} w &= 12. \end{aligned} Find logzw.\log_z w.

Answer: 60
Concepts:logarithmalgebraic manipulation
Difficulty rating: 1930
Small Hint:

Rewrite each given logarithm with base ww

Big Hint:

Expand logw(xyz)\log_w(xyz) as a sum of three logarithms

Solution:

Taking reciprocals of the given logarithms gives logwx=124,logwy=140,logw(xyz)=112. \begin{aligned} \log_w x&=\frac1{24},\\ \log_w y&=\frac1{40},\\ \log_w(xyz)&=\frac1{12}. \end{aligned} Therefore logwz=112124140=160. \log_w z=\frac1{12}-\frac1{24}-\frac1{40} =\frac1{60}. Taking the reciprocal yields logzw=60.\log_z w=60.

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