Q: 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?

A: $$2^a \cdot 2^b\cdot 2^c = 2^{a + b + c} = 2^6\;\;\iff\;\; a + b + c = 6$$
(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 $3$.)

A: Note that
$2^a\cdot2^b\cdot2^c=2^{a+b+c}$ and $64=2^6$.
You know that $2^a\cdot2^b\cdot2^c=64$, then you have that $2^{a+b+c}=2^6$, and from this you conclude that $a+b+c=6$.
Solution $(a,b,c)= (1,2,3)$, therefore $2^a+2^b+2^c= 2^1+2^2+2^3=2+4+8=14$.

A: $$2^{a}\cdot2^{b}\cdot2^{c}=2^{a+b+c}$$
$$a+b+c=6$$

A: Here's the $log$ solution,
$log 2^{a+b+c}=log 64$
$(a+b+c).(0.3010)=6.(0.3010)$
$a+b+c=6$
$a \neq b \neq c \implies log 2^a\neq log 2^b \neq log 2^c$
$log 2^{a+b+c}=log2^a+log2^b+log2^c=6.log2$
$a,b,c$ are distinct, therefore $(a,b,c)=(1,2,3),(3,2,1),(1,3,2),(2,3,1),(2,1,3),(3,1,2)$
Therefore, $2^a+2^b+2^c= 2+ 4+ 8 = 14$
