1976 AMC 12 Problem 20

Attempt Problem 20 of the 1976 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 1976 AMC 12 solutions, or check the answer key.

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20.

Let a,a, bb and xx be positive real numbers distinct from one. Then 4(log⁡ax)2+3(log⁡bx)2=8(log⁡ax)(log⁡bx) \begin{aligned} 4(\log_a x)^2&+3(\log_b x)^2\\ &=8(\log_a x)(\log_b x) \end{aligned}
for all values of a,a, bb and xx
if and only if a=b2a=b^2
if and only if b=a2b=a^2
if and only if x=abx=ab
none of these
Answer: E
Concepts:logarithmquadraticfactoring
Small hint:
Move all terms to one side and factor the quadratic in the two logarithms
Big hint:
Translate each factor equation into a relation between the bases using change of base
Solution:
Factoring gives (2log⁡ax−log⁡bx)⋅(2log⁡ax−3log⁡bx)=0. \begin{aligned} &\bigl(2\log_a x-\log_b x\bigr)\\ &\quad\cdot \bigl(2\log_a x-3\log_b x\bigr)=0. \end{aligned} Since x≠1,x\ne1, change of base shows that the first factor vanishes when a=b2,a=b^2, while the second vanishes when a3=b2.a^3=b^2. Thus the equation holds under either of two conditions, and no listed “if and only if” statement describes their union.
Therefore, the correct answer is E.
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