As part of his project paper, Jim summarized his thoughts on the class question, "At what value of x does 2^x begin to equal 0?"

Jim wrote, "Ramon made a conjecture that 2^x equals 0 for some values of x. I tried using my calculator to solve 2^x = 0 by using logarithms.

Jim continues, "but I got an error message when I tried to find log 0. So, I thought about Ramon's statement that 2^x equals 0 when x equals -100."

"What does 2 ^-100 mean? It means 1 over 2 ¹⁰⁰, or 1 divided by 2 ¹⁰⁰. But, 2 ¹⁰⁰ is a very large number because you start with 2 and double the amount 99 more times. When you divide 1 by a very large number, you get a very small answer. It is such a small answer that a calculator doesn't have enough space to show or work with all the zeros behind the decimal point, so it rounds off the answer to 0. That's what we discovered yesterday.

"So I looked at Ramon's conjecture again: What does 2^x = 0 mean? It means find a power (x) so that if you double 2 enough times, you equal 0. But even if you use a negative number, such as x = -1,000,000, you have a value of 1 over 2 to the millionth power, a very small number. If you then checked x = -1,000,001, you'd get half as much, but not 0. This would continue forever. So, there is not a solution to 2^x = 0. That's probably why I couldn't find a value for log 0. Also, that's why the graph gets very close to the y-axis, but we never see an intersection."

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