# exponent manipulation - $4^{21} \cdot 5^{11} = 2 \cdot 10^{n}$ - what is $n$?

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
16
22
23
24
32

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

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...

Thanks!

Update:

I found the actual question:

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

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You get a star for figuring out what the real question was lol – kralco626 Aug 16 '11 at 2:06
@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. – The Chaz 2.0 Aug 16 '11 at 2:08
Ya, I didn't think to factor the base to get them in the form I wanted. – 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.

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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 – kralco626 Aug 16 '11 at 2:01

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

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can you explain? I'm missing what your saying... – kralco626 Aug 16 '11 at 2:12
$\rm 10^n = (2\cdot 5)^n = 2^n\cdot 5^n\:$ – 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: $$4^{11}5^{21}=(2^2)^{11}5^{21}=2^{22}5^{21}=2^{1} 2^{21}5^{21}=2*10^{21}.$$

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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. – kralco626 Aug 16 '11 at 2:05
is that last part supposed to be 2*10^11 or 2*10^21? – kralco626 Aug 16 '11 at 2:08
It should be ^21. I went through and changed my answers when the question got updated, looks like I missed a spot. – Aaron Aug 16 '11 at 2:30