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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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
Difficulty rating: 1930
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.

Full Exam

Problem 1 in Other Years