2
$\begingroup$

I just recently started relearning math as an adult, this should be easy but I have trouble understanding what the actual question is. I am not just looking for the answer to this, I merely wish to understand what the question is asking.

Express each of the following expressions in the form $2^m3^na^rb^s$, where $m$, $n$, $r$ and $s$ are positive integers.

a) $8a^2b^3(27a^4)(2^5ab)$

$\phantom{\text{ ![a busy cat-miau](http://www3.wolframalpha.com/Calculate/MSP/MSP81a28e31iaea1e9ie00005efb3bgdiecc5cai?MSPStoreType=image/gif&s=15&w=140&h=22)}}$

  • 0
    are you given anywhere that $\gcd(ab,6)=1$, or perhaps that $\gcd(a,b)=1$?2012-07-16
  • 0
    @robjohn: I think you have to assume that, or that you are solving it for general $a,b$.2012-07-16

1 Answers 1

2

You are being asked to express $8$ as $2^3$ and similarly $27$, then to commute the various terms to gather the exponents of $2, 3, a, b$. For example, how many powers of $a$ are in the expression?

  • 0
    7 powers of a? So I get in theory I think what to do, but if I replace 8 with 2 and 27 with 3 but what about the 2^5? Thanks to all who help me!2012-07-16
  • 0
    @nitrous2: Correct. You also need to collect all the factors of $2$, some of which come from the $8$ and some from the $2^5$.2012-07-16
  • 0
    Okay, so plugging things in I am at $2^m3^na^7b^4$ How do I find the 2^m and 3^n?2012-07-16
  • 0
    The answer is $2^83^3a^7b^4$ How would I of found the 2^m and the 3^n?2012-07-16
  • 0
    So 3^3 is a result of factoring 27? 3*3*3?2012-07-16
  • 0
    @nitrous2: Correct. $2^52^3=2^{5+3}=2^8$ and the $3^3$ comes directly from the $27$. These exercises are intended to give practice in handling exponents like this.2012-07-16
  • 0
    Awesome. So the correct solution appears to be: $8a^2b^3(27a^4)(2^5ab) = 2^3a^2b^3(3^3a^4)(2^5*a^1*b^1) = (2^3*2^5)3^3(a^2*a^4*a^1)(b^3*b^1) =2^83^3a^7b^4$2012-07-16
  • 0
    @nitrous2: Your statement is correct, but that isn't the form the solution was requested in. It is a step along the way.2012-07-16
  • 0
    Final Answer: $2^83^3a^7b^4$2012-07-16
  • 0
    @nitrous2: You got it.2012-07-17