1991 AMC 12 Problem 24

Attempt Problem 24 of the 1991 AMC 12 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 1991 AMC 12 solutions, or check the answer key.

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24.

The graph, G,G, of y=log⁡10xy=\log_{10}x is rotated 90∘90^\circ counter-clockwise about the origin to obtain a new graph G′.G'. Which of the following is an equation for G′?G'?
y=log⁡10(x+909)y=\log_{10}\left(\frac{x+90}{9}\right)
y=log⁡x10y=\log_x10
y=1x+1y=\frac1{x+1}
y=10−xy=10^{-x}
y=10xy=10^x
Answer: D
Concepts:graph transformationrotationlogarithmexponential
Small hint:
A 90∘90^\circ counter-clockwise rotation sends (x,y)(x,y) to (−y,x)(-y,x)
Big hint:
Rename the rotated coordinates and solve the logarithmic relation for the new yy
Solution:
A point (u,log⁡10u)(u,\log_{10}u) rotates to (x,y)=(−log⁡10u,u). (x,y)=(-\log_{10}u,u). Hence x=−log⁡10y,x=-\log_{10}y, so log⁡10y=−x\log_{10}y=-x and y=10−x.y=10^{-x}.
Thus the correct answer is D.
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