|
|
| Line 1: |
Line 1: |
| − | == Problem ==
| |
| − | The increasing [[geometric sequence]] <math>x_{0},x_{1},x_{2},\ldots</math> consists entirely of [[integer|integral]] powers of <math>3.</math> Given that
| |
| | | | |
| − | <math>\sum_{n=0}^{7}\log_{3}(x_{n}) = 308</math> and <math>56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,</math>
| |
| − |
| |
| − | find <math>\displaystyle \log_{3}(x_{14}).</math>
| |
| − |
| |
| − | == Solution ==
| |
| − | Suppose that <math>\displaystyle x_0 = a</math>, and that the common [[ratio]] between the terms is <math>r</math>.
| |
| − |
| |
| − |
| |
| − | The first conditions tells us that <math>\displaystyle \log_3 a + \log_3 ar \ldots \log_3 ar^7 = 308</math>. Using the rules of [[logarithm]]s, we can simplify that to <math>\displaystyle \log_3 a^8r^{1 + 2 + \ldots + 7} = 308</math>. Thus, <math>\displaystyle a^8r^{28} = 3^{308}</math>. Since all of the terms of the geometric sequence are integral powers of <math>3</math>, we know that both <math>a</math> and <math>r</math> must be powers of 3. Denote <math>\displaystyle 3^x = a</math> and <math>\displaystyle 3^y = r</math>. We find that <math>8x + 28y = 308</math>. The possible positive integral pairs of <math>(x,y)</math> are <math>(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)</math>.
| |
| − |
| |
| − |
| |
| − | The second condition tells us that <math>56 \le \log_3 (a + ar + \ldots ar^7) \le 57</math>. Using the sum formula for a [[geometric series]] and substituting <math>x</math> and <math>y</math>, this simplifies to <math>3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}</math>. The fractional part <math>\approx \frac{3^{8y}}{3^y} = 3^{7y}</math>. Thus, we need <math>\approx 56 \le x + 7y \le 57</math>. Checking the pairs above, only <math>\displaystyle (21,5)</math> is close.
| |
| − |
| |
| − |
| |
| − | Our solution is therefore <math>\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = 091 \displaystyle</math>.
| |
| | | | |
| | == See also == | | == See also == |
