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In Chapter 1.3, Basic Mathematics, Serge Lang, there is the question:

Express the following expression in the form $2^m3^na^rb^s$, where $m, n, r, s$ are positive integers:$$16b^3a^2(6ab^4)(ab)^3$$

The answer I got was $16 \cdot 6 \cdot a^6 \cdot b^{10}$.

Is there something I did wrong?

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5 Answers 5

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You're on the right track, just not quite finished:

$16*6=2^4 * 2 * 3=2^5 * 3^1$

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  • $\begingroup$ Now I get it. I struggled with that one for a while there.. will +1 when I have enough rep. $\endgroup$
    – usernvk
    May 8, 2013 at 23:30
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What you've done so far is correct... but we need to do a little more...

$$16 \times 6 = 2^4\times 2 \times 3 = 2^5 \times 3$$

So we have that $$16 \cdot 6 \cdot a^6 \cdot b^{10} = 2^5\cdot 3^1\cdot a^6\cdot b^{10}$$

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    $\begingroup$ Deserves a thumbs up! +1 $\endgroup$
    – Amzoti
    May 9, 2013 at 0:31
  • $\begingroup$ Great answer, except the working shows $2^5$ but concluded $2^4$. +1. Also, thanks for the $\cdot$ edit. Couldn't figure out where to find the symbol. $\endgroup$
    – usernvk
    May 9, 2013 at 2:22
  • $\begingroup$ It's just \cdot for formatting "$\cdot$" $\endgroup$
    – amWhy
    May 9, 2013 at 2:27
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Yes, $16\cdot 6$ is not yet in the form $2^m\cdot 3^n$. Otherwise it is correct.

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$6$ isn't a power of $3$. You'll want to change $16\cdot 6$ to $32\cdot 3$.

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Great question, believe it or not, these same questions from that same book stumped me at one point.

This might help you out.

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

The goal here is to reduce everything to it's smallest factors. In these questions they all reduce to smallest factors in the form $2^m3^na^rb^s$.

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  • $\begingroup$ When I couldn't figure out what the exercise was asking, your question was indispensable. That helped me solve (a) but (b) had me stumped. I couldn't figure out how to reduce $16 \cdot 6$ to $2^m \cdot 3^n$. +1 on your original question. $\endgroup$
    – usernvk
    May 9, 2013 at 2:37

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