An algebra problem by Aly Ahmed

Algebra Level 2

3^x+4^x+5^x=6^x


The answer is 3.

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1 solution

We assume that x x is a positive integer (since it is not given).

For all positive integers x x , the unit's place digit of 3 x 3^x can be 1 , 3 , 7 , 9 1,3,7,9 , of 4 x 4^x can be 4 , 6 4,6 , of 5 x 5^x can be 5 5 , and of 6 x 6^x can be 6 6 . So, the unit's place digit of 3 x + 4 x + 5 x 3^x+4^x+5^x has to be 6 6 (since 3 , 5 3,5 are odd and 4 4 is even, the unit's place digit of the L. H. S. can be even only). So, the only possibility is that x = 3 x=\boxed 3 (by direct check).

You don't need to assume that it's an integer at all.

You just need to show that f ( x ) = 3 x + 4 x + 5 x 6 x f(x) = 3^x + 4^x + 5^x - 6^x is a decreasing function. (And add a few more lines... intermediate value theorem.... yadda yadda yadda)

Pi Han Goh - 11 months, 3 weeks ago

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