I was taking a practice GMAT test and it had a question like this:

$4^{21} \cdot 5^{11} = 2 \cdot 10^n$

What is $n$?

The available answers were something like

I'm not exactly sure on the multiple choice options...

However I'm not allowed a calculator so I'm not sure how to go about this...

I'm generally pretty good at the little tricks that allow you to solve a questions without the calculator but I have no idea how to do this one...



I found the actual question:

enter image description here

  • $\begingroup$ Do you mean $4^{11}$ and not $4^{21}$? $\endgroup$ – El'endia Starman Aug 16 '11 at 1:50
  • $\begingroup$ no, that I have seen before, this is deff 21, because I remember the two exponents added up to 32 $\endgroup$ – kralco626 Aug 16 '11 at 1:57
  • $\begingroup$ looks like I mixed up the exp/base combination there though... I posted the real question. $\endgroup$ – kralco626 Aug 16 '11 at 2:02
  • $\begingroup$ Thanks everyone for your answers! They were very prompt and helpful!!! $\endgroup$ – kralco626 Aug 16 '11 at 2:11

I'm assuming you meant $4^{11}\cdot 5^{21} = 2\cdot 10^n$. The rules you want to know for exponents can be found on http://en.wikipedia.org/wiki/Exponentiation (this page has a lot of information, so depending on what you need, you may only be interested in http://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties ). For this question, in particular, you want \begin{align*} (a\cdot b)^n &= a^n\cdot b^n\ \ \operatorname{and}\newline a^{nm} &= (a^n)^m \end{align*}

So looking at $10^n = (2\cdot 5)^n$, we get \begin{align*} 4^{11}\cdot 5^{21} &= (2^2)^{11}\cdot 5^{21}\newline &= 2^{22}\cdot 5^{21}\newline &= 2\cdot (2^{21}\cdot 5^{21}). \end{align*}

So if you question was with the exponents reversed on the 4 and 5, we get that $n=21$.

  • $\begingroup$ You get a star for figuring out what the real question was lol $\endgroup$ – kralco626 Aug 16 '11 at 2:06
  • 2
    $\begingroup$ @kralco: The common thread of all these answers is recognizing the prime factorization of numbers, knowing rules of exponents, and applying the fact that there is equality. $\endgroup$ – The Chaz 2.0 Aug 16 '11 at 2:08
  • $\begingroup$ Ya, I didn't think to factor the base to get them in the form I wanted. $\endgroup$ – kralco626 Aug 16 '11 at 2:11

I hope it was a question like that one, and not exactly that one. $4=2^2$, so $4^{21}=(2^2)^{21}=2^{42}$. $2\times5=10$, so $2^{11}\times5^{11}=10^{11}$. So $$4^{21}\times5^{11}=2^{42}\times5^{11}=2^{31}\times2^{11}\times5^{11}=2^{31}\times10^{11}$$

Trying to write it as $2\times10^n$ for some $n$ will only work if you allow $n$ to be a decimal number, not a whole number.

  • $\begingroup$ I updated my question to include a screen shot of the actual question which I was able to find... thanks for your help so far $\endgroup$ – kralco626 Aug 16 '11 at 2:01

Comparing powers of $5$ in $\rm\:2\cdot 10^n = 5^{21}\cdot 4^{11}\:$ immediately yields $\rm\: n = 21\:.$

  • $\begingroup$ can you explain? I'm missing what your saying... $\endgroup$ – kralco626 Aug 16 '11 at 2:12
  • $\begingroup$ $\rm 10^n = (2\cdot 5)^n = 2^n\cdot 5^n\:$ $\endgroup$ – Bill Dubuque Aug 16 '11 at 2:15

Updating in light of the corrected question. Here are the basic rules that would allow you to solve a problem like this:

  • $(ab)^n=a^nb^n$
  • $(a^n)^m=a^{(mn)}$
  • $a^{m+n}=a^m a^n$

So if we want to analyze $4^{11} * 5^{21}$, we see that we could use the first rule to combine things if we had terms that were all to the same power (which will be the $21$st power), and if we had a $2^{21}$ it would combine with the $5^{21}$ to make a $10^{21}$ (which is a term we want, given that we want to have $10^n$), which motivates the following series of equalities: $$4^{11}5^{21}=(2^2)^{11}5^{21}=2^{22}5^{21}=2^{1} 2^{21}5^{21}=2*10^{21}.$$

  • $\begingroup$ sorry about the question, I was remembering it incorrectly, I was able to go back and find it however, and posted an update in my question. $\endgroup$ – kralco626 Aug 16 '11 at 2:05
  • $\begingroup$ is that last part supposed to be 2*10^11 or 2*10^21? $\endgroup$ – kralco626 Aug 16 '11 at 2:08
  • $\begingroup$ It should be ^21. I went through and changed my answers when the question got updated, looks like I missed a spot. $\endgroup$ – Aaron Aug 16 '11 at 2:30

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