2^4 = 16 AND 4^2 = 16Date: 10/29/2001 at 12:24:38 From: Yvonne Subject: Exponents 2^4 = 16 AND 4^2 = 16. Can you think of any other pair of unequal numbers that share the same relation as 2 and 4 in the above example? What was your strategy? Date: 10/29/2001 at 12:31:14 From: Doctor Ian Subject: Re: Exponents Hi Yvonne, My strategy is to think about symmetry, and the properties of even exponents. Note that 1 (-2)^(-4) = ----- 2^4 What do you suppose (-4)^(-2) is equal to? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 10/29/2001 at 12:51:00 From: Yvonne Ortega Subject: Re: Exponents Thank you very much. I put that down as a possibility, but I thought perhaps there was another pair of numbers I was not thinking of. Are there any other positive numbers you can think of? I sure can't. Thanks again. Date: 10/30/2001 at 11:43:39 From: Doctor Ian Subject: Re: Exponents Hi Yvonne, Let's try another strategy. You can use the University of Vienna's function plotter at http://www.univie.ac.at/future.media/moe/fplotter/fplotter.html to plot the function (2^x) - (x^2) Now, if you do this, any value of x where the graph crosses the x-axis must be a solution to 2^x = x^2 and thus the kind of number we're looking for. If you go ahead and plot this function, you'll see that the graph crosses the x-axis at three points: x = 2 (duh), x = 4 (which we already knew), and at approximately -0.7666. Let's check that: -0.7666^2 = 0.5877 2^-0.7666 = 0.5878 Which is good enough for government work. Now, I know that you were looking for positive numbers... but you were also looking for a strategy, and this looks like a reasonable one to explore. What happens if we try (3^x) - (x^3) We get a graph that crosses the x-axis at two points: x = 3 (duh), and approximately x = 2.478. Let's check: 2.478^3 = 15.216 3^2.478 = 15.216 Now, 2.478 is only an approximation, but you can approximate it as closely as you like by using something like Newton's Method. And you can use the method to find as many pairs of numbers as you like. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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