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2009 AMC 12B Problems/Problem 15

Problem

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

$\textbf{(A)}\ 3(1 + r)^x = 7\qquad \textbf{(B)}\ 3(1 + r/10)^x = 7\qquad \textbf{(C)}\ 3(1 + 2r)^x = 7$ $\textbf{(D)}\ 3(1 + \sqrt {r})^x = 7\qquad \textbf{(E)}\ 3(1 + 1/r)^x = 7$

Solution 1

(B) Intuitively, $x$ will be largest for that option for which the value in the parentheses is smallest.

Formally, first note that each of the values in parentheses is larger than $1$. Now, each of the options is of the form $3f(r)^x = 7$. This can be rewritten as $x\log f(r) = \log\frac 73$. As $f(r)>1$, we have $\log f(r)>0$. Thus $x$ is the largest for the option for which $\log f(r)$ is smallest. And as $\log f(r)$ is an increasing function, this is the option for which $f(r)$ is smallest.

We now get the following easier problem: Given that $0<r<3$, find the smallest value in the set $\{ 1+r, 1+r/10, 1+2r, 1+\sqrt r, 1+1/r\}$.

Clearly $1+r/10$ is smaller than the first and the third option.

We have $r^2 < 10$, dividing both sides by $10r$ we get $r/10 < 1/r$.

And finally, $r/100 < 1$, therefore $r^2/100 < r$, and as both sides are positive, we can take the square root and get $r/10 < \sqrt r$.

Thus the answer is $\boxed{\textbf{(B) } 3\left(1+\frac{r}{10}\right)^x = 7}$.

Solution 2

As stated in the first solution, $x$ will be largest when the value in the parenthesis is smallest.

We can plug in a value for $r$ (for instance, $r=2$) and see which parenthesis has the smallest value. Doing so, we see that B is the answer.

-Made_in_2004

See Also

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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