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kalita
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Exponents Algebraically Updated on: 13 Nov 2015, 10:50
Originally posted by kalita on 13 Nov 2015, 10:37.
Last edited by kalita on 13 Nov 2015, 10:50, edited 3 times in total.
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Hello,

I know it is a silly question which I kind of understand, but not really. Just need some help on this to have it nailed in my head. In this post (see below), Bryan Galvin of Veritas explained exponents.
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/0 ... mment-5735

I asked the same question but it was never addressed and eventually deleted by administrator, probably it is too silly. Hence the question is moving here :) Please clarify how we get from

\(\frac{(2)^{4a}}{2^4}\) to \(a\)? In other words how exactly does the isolating of a happen in this example? I am confused because \((2)^{(4-4)}=2^{0\\
*a}\), and then we get \(a\) to stand on its own, but this does not make sense. If I take \(\frac{2^{4a}}{2^4} = 2^{(4a-4)}\), then \(2^{(4(a-1))}\)...I don't know, I am lost doing this algebraically/mechanically. There is a rule I am missing and I don't see what it is. Can you help me break this down in pure algebra?

Eventually I explained it to myself this way: since \(a\) is some kind of number representing another number of \(2\)s, then \(2^{4a}\) essentially means \(2*2*2*2*a\) (where \(a\) is few other \(2\)s) and when you divide by \(2^{4}\) or by \(2*2*2*2\), you are indeed left with \(a\). But my challenge is that algebraically I am confusing myself and I can't get to the same result/logic.

Thanks for help.



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Exponents Algebraically 13 Nov 2015, 10:46
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Hello,

I know it is a silly question which I kind of understand, but not really. Just need some help on this to have it nailed in my head. In this post (see below), Bryan Galvin of Veritas explained exponents.
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/02 ... mment-5735

I asked the same question but it was never addressed and eventually deleted by administrator, probably it is too silly. Hence the question is moving here :) Please clarify how we get from

\(\frac{(2)^{4a}}{2^4}\) to \(a\)? In other words how exactly does the isolating of a happen in this example? I am confused because \((2)^{(4-4)}=2^{0\\
*a}\), and then we get a to stand on its own, but this does not make sense. If I take \(\frac{2^{4a}}{2^4} = 2^{(4a-4)}\), then \(2^{(4(a-1))}\)...I don't know, I am lost doing this algebraically/mechanically. There is a rule I am missing and I don't see what it is. Can you help me break this down in pure algebra?

Eventually I explained it to myself this way: since \(a\) is some kind of number representing another number of 2s, then 2^{4a} essentially means 2*2*2*2*a (where a is few other 2s) and when you divide by 2^4 or by 2*2*2*2, you are indeed left with \(a\). But my challenge is that algebraically I am confusing myself and I can't get to the same result/logic.

Thanks for help.
Show more

Your question might have been deleted as you did not follow the proper posting guidelines (link in my signatures). I have reformatted your question as it was not getting displayed properly and with exponents be careful about how your post is getting displayed.

As for your question,

\(\frac{2^{4a}}{2^4}\) = \(2^{4a}*2^{-4}\) = \(2^{4a-4}\) ... based on the fact that \(\frac{1}{a} = a^{-1}\) and \(a^m*a^n = a^{(m+n)}\)

As for your own method of explanation, it is correct to assume that a = any number (integer or otherwise) and that if lets say a=1, then 4a= 4 ---> \(2^{4a}\) = \(2^4\) = \(2*2*2*2 = 16\)

Hope this helps.
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Exponents Algebraically 13 Nov 2015, 11:11
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kalita
Hello,

I know it is a silly question which I kind of understand, but not really. Just need some help on this to have it nailed in my head. In this post (see below), Bryan Galvin of Veritas explained exponents.
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/02 ... mment-5735

I asked the same question but it was never addressed and eventually deleted by administrator, probably it is too silly. Hence the question is moving here :) Please clarify how we get from

\(\frac{(2)^{4a}}{2^4}\) to \(a\)? In other words how exactly does the isolating of a happen in this example? I am confused because \((2)^{(4-4)}=2^{0\\
*a}\), and then we get a to stand on its own, but this does not make sense. If I take \(\frac{2^{4a}}{2^4} = 2^{(4a-4)}\), then \(2^{(4(a-1))}\)...I don't know, I am lost doing this algebraically/mechanically. There is a rule I am missing and I don't see what it is. Can you help me break this down in pure algebra?

Eventually I explained it to myself this way: since \(a\) is some kind of number representing another number of 2s, then 2^{4a} essentially means 2*2*2*2*a (where a is few other 2s) and when you divide by 2^4 or by 2*2*2*2, you are indeed left with \(a\). But my challenge is that algebraically I am confusing myself and I can't get to the same result/logic.

Thanks for help.

Your question might have been deleted as you did not follow the proper posting guidelines (link in my signatures). I have reformatted your question as it was not getting displayed properly and with exponents be careful about how your post is getting displayed.

As for your question,

\(\frac{2^{4a}}{2^4}\) = \(2^{4a}*2^{-4}\) = \(2^{4a-4}\) ... based on the fact that \(\frac{1}{a} = a^{-1}\) and \(a^m*a^n = a^{(m+n)}\)

As for your own method of explanation, it is correct to assume that a = any number (integer or otherwise) and that if lets say a=1, then 4a= 4 ---> \(2^{4a}\) = \(2^4\) = \(2*2*2*2 = 16\)

Hope this helps.
Show more

Thank you for your effort, I am still not clear as to how after \(2^{4a}*2^{-4}\) = \(2^{4a-4}\) (that I understand really well actually) I am left with just \(a\) please? How is \(a\) being isolated from \(2^{4a-4}\) (what are the next steps please) or back to my original question what are the exact algebraic steps that lead me from \(\frac{2^{4a}}{2^4}\) to just \(a\), being in exponent in the beginning and becoming a standalone "base"/number. Sorry if it is repetitive but I don't seem to get the answer from your explanation. And apologies for kindergarten but it is necessary apparently.

Even in this post, Bryan simply jumped from \(2^{4a}\) to \(a\) as if it is \(2+2\). It probably is, just not for me. Thanks for confirming my common sense.

P.S I did my best to format my post well this time!
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Exponents Algebraically 13 Nov 2015, 11:24
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kalita
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kalita
Hello,

I know it is a silly question which I kind of understand, but not really. Just need some help on this to have it nailed in my head. In this post (see below), Bryan Galvin of Veritas explained exponents.
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/02 ... mment-5735

I asked the same question but it was never addressed and eventually deleted by administrator, probably it is too silly. Hence the question is moving here :) Please clarify how we get from

\(\frac{(2)^{4a}}{2^4}\) to \(a\)? In other words how exactly does the isolating of a happen in this example? I am confused because \((2)^{(4-4)}=2^{0\\
*a}\), and then we get a to stand on its own, but this does not make sense. If I take \(\frac{2^{4a}}{2^4} = 2^{(4a-4)}\), then \(2^{(4(a-1))}\)...I don't know, I am lost doing this algebraically/mechanically. There is a rule I am missing and I don't see what it is. Can you help me break this down in pure algebra?

Eventually I explained it to myself this way: since \(a\) is some kind of number representing another number of 2s, then 2^{4a} essentially means 2*2*2*2*a (where a is few other 2s) and when you divide by 2^4 or by 2*2*2*2, you are indeed left with \(a\). But my challenge is that algebraically I am confusing myself and I can't get to the same result/logic.

Thanks for help.

Your question might have been deleted as you did not follow the proper posting guidelines (link in my signatures). I have reformatted your question as it was not getting displayed properly and with exponents be careful about how your post is getting displayed.

As for your question,

\(\frac{2^{4a}}{2^4}\) = \(2^{4a}*2^{-4}\) = \(2^{4a-4}\) ... based on the fact that \(\frac{1}{a} = a^{-1}\) and \(a^m*a^n = a^{(m+n)}\)

As for your own method of explanation, it is correct to assume that a = any number (integer or otherwise) and that if lets say a=1, then 4a= 4 ---> \(2^{4a}\) = \(2^4\) = \(2*2*2*2 = 16\)

Hope this helps.

Thank you for your effort, I am still not clear as to how after \(2^{4a}*2^{-4}\) = \(2^{4a-4}\) (that I understand really well actually) I am left with just \(a\) please? How is \(a\) being isolated from \(2^{4a-4}\) (what are the next steps please) or back to my original question what are the exact algebraic steps that lead me from \(\frac{2^{4a}}{2^4}\) to just \(a\), being in exponent in the beginning and becoming a standalone "base"/number. Sorry if it is repetitive but I don't seem to get the answer from your explanation. And apologies for kindergarten but it is necessary apparently.

Even in this post, Bryan simply jumped from \(2^{4a}\) to \(a\) as if it is \(2+2\). It probably is, just not for me. Thanks for confirming my common sense.

P.S I did my best to format my post well this time!
Show more

Ok, now I understand your question correctly and can clearly see why you are having that particular question. The original post has a typo.

It is mentioning 16^a=32^b when in the explanation below, it says "Here there’s only one exponential term, 32 to the b power". So technically, the question is:

16*a=32^b ---> \((2^4)*a = (2^5)^b\) ---> \((2^4)*a = (2^5)^b\) ---> dividing both sides by \(2^4\) you get , \(a = \frac{ (2^5)^b}{2^4} = 2^{5b-4}\) and thus C is the correct answer.

Hope this helps.

P.S. You could have posted this question as a new question in the PS forum as well.
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Exponents Algebraically 13 Nov 2015, 13:25
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It makes complete sense now. This is unbelievable. I have tried to solve the "mystery" for so long when it has been about a typo all along. This is nuts, but thanks to you I can now proceed to a peaceful sleep! I would post it in the PS forum, I was just very preoccupied with my exponents skills and this thing has been a nightmare quite honestly. And this is coming from a 45Quant (not that this is a lot) guy :) I mean I was so focused on this one technical thing that everything else, such as "Here there’s only one exponential term" or "\(2^4 * a = (2^5)^b\)" did not matter.

Now, I am interested, for the curiosity sake, can we isolate \(a\) if hypothetically there were no typo. In other words, can we express \(a\) in terms of \(b\) if we have "original" condition not as intended, but as \(16^{a}=32^{b}\), or bring \(a\) from being an exponent down to representing just a number? Magically, I have been able to do/explain it (as per above) in a non-algebraic way, so there must be a way to do it algebraically. In other words, is this, \(2^{4a−4}\), as far as we can go if we wanted to isolate \(a\) or what direction would we take? I want to get rid of \(2^{4}\) from \(2^{4a}\) expression, to have \(a\) as a base, is this possible? Perhaps you have to use logs or something.

Anyhow, don't bother if it is too difficult. It is definitely beyond the scope.

It is a good day, thank you so much. I will repost the question as appropriate later.

P.S. GMATTERs beware of inattentive reading :) Geez.
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Exponents Algebraically 13 Nov 2015, 13:35
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kalita
It makes complete sense now. This is unbelievable. I have tried to solve the "mystery" for so long when it has been about a typo all along. This is nuts, but thanks to you I can now proceed to a peaceful sleep! I would post it in the PS forum, I was just very preoccupied with my exponents skills and this thing has been a nightmare quite honestly. And this is coming from a 45Quant (not that this is a lot) guy :) I mean I was so focused on this one technical thing that everything else, such as "Here there’s only one exponential term" or "\(2^4 * a = (2^5)^b\)" did not matter.

Now, I am interested, for the curiosity sake, can we isolate \(a\) if hypothetically there were no typo. In other words, can we express \(a\) in terms of \(b\) if we have "original" condition not as intended, but as \(16^{a}=32^{b}\), or bring \(a\) from being an exponent down to representing just a number? Magically, I have been able to do/explain it (as per above) in a non-algebraic way, so there must be a way to do it algebraically. In other words, is this, \(2^{4a−4}\), as far as we can go if we wanted to isolate \(a\) or what direction would we take? I want to get rid of \(2^{4}\) from \(2^{4a}\) expression, to have \(a\) as a base, is this possible? Perhaps you have to use logs or something.

Anyhow, don't bother if it is too difficult. It is definitely beyond the scope.

It is a good day, thank you so much. I will repost the question as appropriate later.

P.S. GMATTERs beware of inattentive reading :) Geez.
Show more


Attentive reading is a very crucial skill for GMAT. Posting a question in the PS forum will generate some good, useful discussion from experts and members alike wherein you will end up learning a lot more.

You are adopting the correct approach with how to deal with bringing a to the based but without an actual question, it is very difficult to judge which method is better. GMAT Quant is more about picking the best strategy in terms of application and time management. You can have many different ways to solve 1 question but the main thing to evaluate here will be whether you will be able to save any time with your choice of method.

If you have \(16^a = 32^b\) ---> \(2^{4a} = 2^{5b}\) ---> 4a = 5b ---> a= 5b/4. This is one way of expressing 'a' as a particular number.

If you are asking how to bring \(2^{4a-4}\) to something \(a = xyz\), then logs will be the way forward. As GMAT does not require knowledge of logs, I will say not to spend too much time on this.

Hope this helps.

P.S. if you do end up with a question such as

\(2^{4a-4} = k\) , where k is any number,

then in order to calculate a in terms of k, take log (to the base e) on both sides,

(4a-4)*log 2 = log k ---> 4a-4 = log k / log 2 ---> a = 0.25* (4+log k/log 2) .



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