2009 AMC 12B Problem 15

Below is the professionally curated solution for Problem 15 of the 2009 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

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Concepts:logarithm inequality

Difficulty rating: 1710

15.

Assume $0 \lt r \lt 3.$ Below are five equations for $x.$ Which equation has the largest solution $x?$

$3(1 + r)^x = 7$

$3(1 + r/10)^x = 7$

$3(1 + 2r)^x = 7$

$3(1 + \sqrt{r})^x = 7$

$3(1 + 1/r)^x = 7$

Solution:

Each equation gives $x = \dfrac{\log(7/3)}{\log(1 + f(r))},$ which is largest when the positive quantity $f(r)$ is smallest.

For $0 \lt r \lt 3,$ among $r,\ \dfrac{r}{10},\ 2r,\ \sqrt{r},\ \dfrac{1}{r},$ the smallest is $\dfrac{r}{10}$ : it is below $r$ and below $\sqrt{r}$ since $r \lt 3 \lt 100.$ So equation (B) has the largest solution.

Thus, the correct answer is B.

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