Question

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

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

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

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

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

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

Answer: B

Concepts:logarithm inequality

Difficulty rating: 1710

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 $0 \lt r \lt 3,$ we have $\dfrac r{10}\lt r\lt2r$ and $\dfrac r{10}\lt\sqrt r$ because $r\lt100.$ Also $r^2\lt9\lt10,$ so $\dfrac r{10}\lt\dfrac1r.$ Thus $r/10$ is the smallest of the five quantities, and equation (B) has the largest solution.

Thus, the correct answer is B.

2009 AMC 12B Problem 15

Answer: B

Difficulty rating: 1710

Solution:

Problem 15 in Other Years