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?

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, $2^a+2^b+2^c= 2+ 4+ 8 = 14$
