2009 AMC 12B Problems/Problem 15

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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

Intuitivelly, $x$ 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 $\boxed{\text{(B) } 3(1 + r/10)^x = 7}$ .

2009 AMC 12B (Problems • Resources)
Preceded by Problem 14	Followed by Problem 16
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2009 AMC 12B Problems/Problem 15

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