<s>[INST] Variable Exponents Question If $a,b$ and $c$ are different positive integers and $2^a\cdot2^b\cdot2^c =64$ then  $2^a+2^b+2^c$=?
This is so far my work: I got $2^a\cdot2^b\cdot2^c=2^6$ then $abc=6$ is this so far in the right track?
 [/INST] $$2^a \cdot 2^b\cdot 2^c = 2^{a + b + c} = 2^6\;\;\iff\;\; a + b + c = 6$$
(Recall that $2^x\cdot 2^y = 2^{x+y}.$)
The only possible combinations of distinct $a, b, c$ which sum to $6$ is $\;(a, b, c) = (1, 2, 3),\;$ or any permutation thereof.*
That is, $$a \neq b \neq c \implies 2^a + 2^b + 2^c = 2^1 + 2^2 + 2^3 = 2 + 4 + 8 = 14.$$
*(Because of symmetry, we do not need to consider $a = 2, b = 1, c = 3,$ etc. All we need to know is if a, b, c are distinct, then we need one of a, b, c to be $1$, one to be $2$, and one to be $3$.)
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