## Sunday, January 26, 2014

### 8: More Exponents

From Dan Meyer (via email):

Which is bigger: \$60^{63}\$ or \$63^{60}\$?

The beauty of this problem is that both sides are beyond the capability of a TI-84 so a reduction of some kind is called for. wolframalpha.com has no difficulty but the answer is startling.

Find Dan on Twitter @ddmeyer, too.

1. Wow, answer on Wolframalpha was pretty surprising!
Is it true for all numbers a and b, when a < b, then a^b > b^a ?

1. That sounds like a good question for Day 29. Thank you!

2. This one really has me thinking. How could a middle school or high school kid reason through this and come up with an answer. Not sure it is possible to know for sure without binomial expansion.
I started by using laws of exponents and prime factoring to divide out what they had in common (Not much between 60 & 63). Then I started considering other pairs of numbers (60 & 64, or 15 & 30). But every time I'm able to reduce some but am still left with a "now what?" statement. I know I could use binomial expansion, but not sure what other approaches to take. I really like the open ended-ness of this and that even an Alg2 kiddo could be thinking about number sense and connecting that number sense back to concepts like polynomial factoring to consider which might be larger.

Thanks for the deep thought and diversion!

3. Are (scientific) calculators allowed? If so, just calculate (60/63)^60 times (60)^3, which is (60)^63 divided by (63)^(60), and you'll find this quotient is greater than 1, which implies that (60)^(63) is greater than (63)^(60). Without a calculator (and without calculus), the only thing I can think of is that (63)^(60) divided by (60)^(63) equals B^3 divided by (60)^3, where B = (63/60)^(20) = (21/20)^(20) = (1 + 1/20)^(20). Since B is approximately equal to Euler's constant e = 2.71828... (recall continuous compounding from precalculus), this division is approximately e^3 divided by (60)^3, which in turn is equal to (e/60)^3, which is clearly less than 1.

4. Is it true for all numbers a and b, when a < b, then a^b > b^a ?
As long as you're interested in more than integer-only solutions, ask Wolfram|Alpha to graph x^y = y^x. Also, check out the great intersection point on the graph.

5. Regarding when a^b is greater than b^a, see http://math.stackexchange.com/questions/517555/fastest-way-to-check-if-xy-yx