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 log⁡xw=24,log⁡yw=40,log⁡xyzw=12. \begin{aligned} \log_x w &= 24,\\ \log_y w &= 40,\\ \log_{xyz} w &= 12. \end{aligned} Find log⁡zw.\log_z w.
Answer: 60
Concepts:logarithmalgebraic manipulation
Small hint:
Rewrite each given logarithm with base ww
Big hint:
Expand log⁡w(xyz)\log_w(xyz) as a sum of three logarithms
Solution:
Taking reciprocals of the given logarithms gives log⁡wx=124,log⁡wy=140,log⁡w(xyz)=112. \begin{aligned} \log_w x&=\frac1{24},\\ \log_w y&=\frac1{40},\\ \log_w(xyz)&=\frac1{12}. \end{aligned} Therefore log⁡wz=112−124−140=160. \log_w z=\frac1{12}-\frac1{24}-\frac1{40} =\frac1{60}. Taking the reciprocal yields log⁡zw=60.\log_z w=60.
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