Question 1209405
<br>
{{{ab^a=648=(2^3)(3^4)}}}<br>
The only prime factors of 648 are 2 and 3.<br>
a is an exponent in the expression; the exponents in the prime factorization of 648 are 3 and 4.  That means a can only be 1, 2, or 3.<br>
(1) a=1<br>
{{{(1)(b^1)=648}}}
{{{b=648}}}<br>
first solution: (a,b) = (1,648)<br>
(2) a=2<br>
{{{(2)(b^2)=648}}}
{{{b^2=648/2=324}}}
{{{b=18}}} or {{{b=-18}}}<br>
second and third solutions: (a,b) = (2,18) and (a,b) = (2,-18)<br>
(3) a=3<br>
{{{(3)(b^3)=648}}}
{{{b^3=216}}}
{{{b=6}}}<br>
fourth solution: (a,b) = (3,6)<br>
ANSWER: 4 pairs of integer (a,b) satisfy the equation ab^a = 648<br>