<p>pg. 549 #7... </p>
<p>(2^a) x (2^b) x (2^c) = 64, then (2^a)+(2^b)+(2^c) =</p>
<p>I solved it by plugging in random integers...is there a faster way?</p>
<p>(2^a) x (2^b) x (2^c) = 64 is also, if you factor out two:</p>
<p>2^(a+b+c) equals (2^a)+(2^b)+(2^c)</p>
<p>Because (2^a) x (2^b) x (2^c), 2^(a+b+c), and (2^a)+(2^b)+(2^c), are equal, they all equal 64.</p>
<p>I haven't taken the SAT for a year, so I'm a little rusty.</p>
<p>^They're not all equal. :| </p>
<p>2^a x 2^b x 2^c = 2^(a+b+c) = 64 = 2^6. </p>
<p>a + b + c = 6. </p>
<p>Unless the question specifies further (such as saying a, b, and c are distinct positive integers), any three numbers which add to 6 can be subbed in for a, b, and c, giving you infinite possible answers.</p>