GMAT Ninja Quant Ep 3: Exponents

[@All_Walks_]

I'm not going to lie, this might have been the hardest video in the series

[@emiliemarcello8301]

This video was fucking nuts

[@Spencer-oi5xx]

Please tell me this isn't worse than properties of numbers...

[@Alappavan]

Nah bro it's the simplest

[@aj1218]

Did you get into the school you wanted?

[@lucianopavarotti-yq3pf]

Good lol I am not alone

[@rejoicingGrace]

Some are among the hardest exponent questions that I saw recently but they come with helpful tips and efficient solutions. Really appreciate this session and look forward to next one! Thank you GMAT Ninja team!

[@GMATNinjaTutoring]

Thank you so much, Grace! Always lovely having you join the premieres. Have fun studying!

[@sims_ran]

Not gonna lie, you had us in the first half!

[@arpitbafna8265]

Can you please mark the answers, that makes it really convenient for when I'm just cross-checking :)

[@bombaybarbell]

Absolutely amazed by the way you approach solving these problems! Can't thank enough. Loved it.

[@GMATNinjaTutoring]

Thank you so much, Abhishek. Have fun studying!

[@jheteale]

Hey Bransen, thanks for the vid! In Q8. Could you square a^b/2 > a^2b, to show a result of a^b > a^4b. This would arrive at the same conclusion, that a

[@prachitabakliwal5207]

Hi Bransen, in the second last question, I solved using following technique. Q : a^b/2> is b/2> b

[@shirsendumaiti5682]

i think you reading the question wrong it's a^b/2 > a^2b not '

[@prajakta3921]

@@shirsendumaiti5682 would this be a valid approach? a^b/2>a^2b, after squaring both sides- a^b>a^4b subsequently, b>4b Options- 1. when 1 1^4x2 which means 1>1 but 1=1 so it does not answer our question, as there exists a scenario where we are getting the sides as equal, and not greater than or less than.

[@ashishsinha9035]

Thanks GMAT Ninja Tutoring for sharing and solving some of the toughest questions on Exponents !

[@GMATNinjaTutoring]

Thank you for watching and suffering through them! :)

[@seanturk2261]

Probably need to expand more on how you got the answer for number 4 skipping through the factoring part helps no one understand how you were able to move the 1 into the exponent and know that it was positive. I understand basic exponential rules but that one is a mind twister and I would be surprised if anyone my level even understands what is going on.

[@HalfGermany100]

Best of luck all of you. Pretty challenging video tbh

[@NeuralNewsletters]

TLDR: Understanding and applying exponent rules is crucial for solving challenging math problems on the GMAT, and practicing with advanced exponent questions can improve efficiency and flexibility in solving these types of problems. 00:00 📈 This video covers challenging quant questions on exponents and quadratics, focusing on advanced topics and providing practice questions for those looking to improve their efficiency and flexibility in applying exponent rules. 11:03 📝 Simplify exponents by manipulating and applying exponent rules to eliminate answer choices and solve the problem step by step. 17:30 📝 Understanding exponent rules and simplifying complex expressions is crucial for solving challenging math problems on the GMAT. 29:54 📝 Understand exponents, factor out lesser exponent, simplify equations, find values of a and b, and solve challenging questions together. 44:05 📝 Exponent questions on the GMAT provide valuable skills and takeaways, including finding factors of numbers and checking for divisibility by odd numbers. 52:51 📝 The speaker discusses how to simplify and compare exponents in a GMAT question, emphasizing the importance of logical deduction and statement sufficiency. 01:05:40 📝 Rewrite equations in scientific notation, manipulate exponents, and cancel out terms to simplify expressions in GMAT quant questions. 01:13:18 📈 Remember key strategies for solving exponent problems and don't worry if you struggled with challenging questions, as you can still achieve a good quant score.

[@ritik2044]

thank you for this - i was making handwritten notes about the exact to come back to this video just before the exam, but your comment has helped me with it much faster

[@ryantash4223]

Great video Bransen, tough but engaging 👍

[@user-gb2tu6fg2g]

I couldn't get my head around the last question at all. But I noticed something - if we divide the last digit of the numerator by the last digit of the denominator for both numbers, we get 9/1=9 and (7)5/5=(1)5. This means that whatever the first number is, its last digit on the right from the decimal point must be 9 and the last digit of the second number must be 5. When we substract number 1 and number 2 we get ......9 - .......5 = .......4 and the answer is the .994. Could somebody tell me if this is a valid way to solve this question?

[@harshmalik3470]

This works when you have one of the options with unit digit as 4, what is there are three of them?

[@TheAmigoBoyz]

wow number 9 is actually crazy -I wonder how anyone would ever be able to do that in 2 minutes during a GMAT

[@GMATNinjaTutoring]

Yeah, that's why we put it towards the end of the video -- it's about as hard as GMAT exponents questions get. Is it possible to answer something like that in around two minutes? Yes, but you DEFINITELY don't need to be that skilled or fast to get an amazing score on the GMAT quant section. :) Have fun studying!

[@SosoCaramello]

thank you so much for these videos!!!!! doing the lords work with these run downs

[@GMATNinjaTutoring]

Thank you so much for the kind words, and have fun studying!

[@MattProchaska]

that was absolutely brutal lol, got a lot of work to do

[@jaywiener5667]

On the last question I think this is a better way to think about it. Start with estimation and think, is 8.99/3.01 bigger or smaller than 3, and same with the 3.99/ 1.995 in terms of 2, and then its apparent the answer should be less than 1. (8.9 / 3.1 is smaller than 3, and 3.99 / 1.9 is slightly bigger than 2 ) as estimates. Then I thought about the terms logically and have this trick.. think about 3.99975 / 1.995 --> .000025 / .005 . (The difference from 4/2. and you can move the decimal point and quickly find its .025/ 5 = .05. Same thing with the other part. Think about 8.999999/ 3.001 --> as .000001 / .001 you can move the decimal and have .001/1 = .001. .005 plus .001 equals the .06 difference from 1. The only drawback is you have to understand if the estimates are greater or lower than the actual values of both terms. And you have consider if you should add that .001 or subtract it from the .005. Since the 3.99/ 1.9 term is slightly above 2, I knew to add it, but if it was actually smaller than 2 in my estimate from the beginning, then you would have to subtract it.

[@antonpeleah3040]

I dont understand why in question 5 when we factor out the 5^a-2 we get the first value in the brackets to 5^2. I am also confused about factoring in the 4^5 + ... question. I thought you could make it (4*4)^5.

[@somasundaram5744]

in Q4 (1/-4(^4a)) and (1/-2(^8a)) is obviously positive, so we can eliminate some options. Then taking some terms common we can just arrive to option D wryt?

[@tdavis5284]

holy moly, math so hard. Thank you for the help!

[@GMATNinjaTutoring]

Thank you for watching! The math feels easier with time and effort -- I promise. :)

[@149prajeeshps5]

I haven't learned exponents yet, but still I managed to get 7/9 correct, (the last one seemed extremely new to me) does that mean I am good with exponent Questions.

[@marcopecorale2221]

Hi, in the question 5, can I factor out also for 5^a ?

[@dushyantkanal8675]

Hi Bransen Quick question on 4th one I follow when you say that a negative number raised to an even exponent will lead to a positive number but here we haven't yet executed the power right? What I mean is, for eg -2^8 = +256 only after the exponents have been used for the repeated multiplication. Similarly, when we have variables, shouldn't the negative hold true for the base till the time the exponents have been utilized an even number of times to then neutralize the negative? Drawing from the above shouldn't the answer actually be option A and not D? Or can it be either option?

[@GMATNinjaTutoring]

Hi Dushyant! It may help to think about another example. Consider: (-2^3)^4. In the video, we're saying that you can disregard the negative because it is eventually raised to an even power anyway. So, (-2^3)^4 = (2^3)^4 = (-8)^4 = (8)^4 = 2^12. Does that make sense?

[@dushyantkanal8675]

I follow that, hence, with that logic, in the question, we can say option A=option D, right?

[@GMATNinjaTutoring]

The problem with (A) is that I end up with a negative result (if I were to plug something in for the variable a in the answer choice). I know that I won’t get a negative result from the initial prompt because everything is raised to an even power. (D), on the other hand, accounts for the fact that everything is raised to a positive power, so I should have 2 raised to an exponent instead of -2 raised to an exponent.

[@harshikadeorah3912]

Hey I don't get this logic as well, can you help understand why should our approach be to change the sign of the base? why should I change the sign of the base when an exponent is still there and write ( -2)^8a = 2^8a when the first one gets me to an answer which is one of the answer choices @@GMATNinjaTutoring

[@srilanka739]

q8 very solvable using test cases. algebra to get to where you did seems tough :(

[@vikramahlawat4933]

My fourth question answer is A Could you explain the last step again

[@JEAND-kl6fo]

For the seventh question I took another path where I had a look at answer choices onlly. I looked for the one that did not belong ---> 11! (because 39 = 3*13 and 91 = 13*7 which means 3 and 7 should belong to S)

[@prasannakumar6742]

Partly true. Yeah if 39 or 91 is present in S then 3 and 7 should also be in S. But the inverse isn't true. There is a possibility that n is divisible by 3 or 7 but not 39 or 81.

[@shermainn]

Hi Branson, just curious why question 8 was not answer E, as a1 ? Would you be able to elaborate more on this, thanks!

[@GMATNinjaTutoring]

The process Bransen demonstrated at the start of the solution was about rephrasing the question, not about finding a solution. This process showed that we could rephrase the question from "is a^(b/2) > a^2b?" to "is a < 1?". At this point, we don't know that a < 1, we've rephrased the question so that we want to know whether the information given in statement (1) or (2) is sufficient to tell us whether a < 1. The information in statement (1) tells us that a > 1, so we know for sure that a is not less than 1. This means the information in statement (1) is sufficient to answer the question. Since we can show the information in statement (2) is not sufficient to tell whether a < 1, the answer to this question is (A). I hope that helps!

[@shermainn]

@@GMATNinjaTutoring This helps alot, Thanks!

[@benjaminng5078]

For qn8, when proving that statement one is sufficient, you deduced that for b^a > b , a has to be greater than 1. However, if a is 0.25 and b is 0.5 then wouldnt it still equate to b^a > b?

[@GMATNinjaTutoring]

From statement (1), we know two things. We know that b^a > b, but we ALSO know that b > 1. This means we can't use a = 0.25 and b = 0.5 as an example scenario for statement (1) as b must be greater than 1. I hope that helps!

[@prajakta3921]

can someone please help me understand if my approach to the 8th question was correct- a^b/2>a^2b, after squaring both sides- a^b>a^4b subsequently, b>4b Options- 1. when 1 1^4x2 which means 1>1 but 1=1 so it does not answer our question, as there exists a scenario where we are getting the sides as equal, and not greater than or less than.

[@GMATNinjaTutoring]

There's a lot of great reasoning in your approach! Unfortunately, there's one small problem in the step where you say a^b > a^4b and, therefore, b > 4b. This step works if a > 1. For example, if a = 2 and so 2^b > 2^4b, then we can definitely say that b > 4b. However, this step doesn't work if 0 < a < 1. For example, if a = 1/2 and so (1/2)^b > (1/2)^4b, then b < 4b. At this point in the solution, we don't know whether 0 < a < 1 or 1 < a, so we can't say for certain whether the step from a^b > a^4b to b > 4b will work. I hope that helps, and I'm sorry your method didn't work out as it's an innovative and creative way of looking at this question!

[@prajakta3921]

@@GMATNinjaTutoring whoa appreciate the quick reply! I was going positively mental over this question and whether my approach is technically sound or not. A bit disappointed to know it's not, but I did know from the start I'll have to work harder on my exponents. Thank you and love this series!!

[@nehashashidhar462]

Actual I'm also following the same method. And it provides appropriate answer for many questions. I've noticed this in GMAT Focus edition book also. So, I guess it's ok to move on with this method.

[@prajakta3921]

@@nehashashidhar462 great, would you like to connect to discuss GMAT related doubts/queries possibly? Idk anyone else who's prepping for the same😅

[@harshmalik3470]

@@prajakta3921 Have you given the exam, how was it?

[@dovidsafir7085]

On question 3 - the bases were equal (4) but the exponents weren't equal? Since they're all base 4, shouldn't 5+5+5+5 = 3x? But that gets the wrong number? What am I doing wrong here? does this only work when there's only 1 term left on each side?

[@GMATNinjaTutoring]

Hi! The problem is you cannot equate the exponents until you've got one and only one term of each base on each side of the equation. We can equate the exponents for 2^3p = 2^9, and we can equate the exponents for (2^4)(3^a) = (2^b)(3^5) because there's only one term for each prime base on each side of the equation. We can't equate the exponents for 4^5 + 4^5 + 4^5 + 4^5 = 64^x because there are multiple terms with a base of 4 on the left-hand side. To see how to approach this problem, consider the sum y + y + y + y. Instead of summing these terms, we could turn it into a multiplication and say y + y + y + y = 4(y). The important thing here is that there are 4 of the same thing. In question 3, we have 4^5 + 4^5 + 4^5 + 4^5. Here too, we have 4 of the same thing so in exactly the same way we said y + y + y + y = 4(y), we can say that 4^5 + 4^5 + 4^5 + 4^5 = 4(4^5). From there, we can say that 4(4^5) = 4*(4^5) = (4^1)*(4^5) = 4^(5+1) = 4^6 We can treat the right-hand side of this question separately and say 64^x = (4^3)^x = 4^3x. Setting these two equal to each other, we go from 4^5 + 4^5 + 4^5 + 4^5 = 64^x to 4^6 = 4^3x. At this point, we can equate the exponents to get 6 = 3x which means x = 2. I hope that helps!

[@Ssrssj]

Hello Bransen, solving without converting to fractions, 4th question's solution would be -2^-8a + -2^-8a = 2(-2^8a) = -2^8a+1 right? Wouldn't the answer choice be A?

[@GMATNinjaTutoring]

Hi Ram! Part of the problem is that you dropped the negative in the exponent. It should be that -2^-8a + -2^-8a = 2(-2^-8a). From here, we know that the -2 is raised to an even power, so I can drop the negative in the base and say that I have 2(2^-8a) = 2^(-8a + 1), which is (D). I hope that helps!

[@Ssrssj]

@@GMATNinjaTutoring Understood! Thanks for clarifying :)

[@vedanth19]

Can we solve Q7 like this? I felt it was much easier and got me the answer quicker before I saw the explanation. n = 5^6 - 2^6 n = (5^2 - 2^2)^3 n = [(5+2)(5-2)]^3 n = [(7)(3)]^3 So we know 7 and 3 are factors, and we can then find the odd one out from the options. Looking forward to your response. I'm loving this series btw, super helpful! Thanks a ton!

[@GMATNinjaTutoring]

Good question! Notice that 5^6 - 2^6 is not actually equal to (5^2 - 2^2)^3. Why is that? To see that, let's consider a more generic example: (a - b)^3 = (a -b)(a- b)(a -b) = a^3 - 3a^2b + 3ab^2 - b^3. In other words, (a - b)^3 does NOT equal (a^3 - b^3), because you get a whole bunch of middle terms when you distribute. For that same reason, (5^2 - 2^2)^3 does NOT equal 5^6 - 2^6. I hope that helps!

[@devamshidinesh4132]

i do not understand how for q5 , the last step , you paired the exponents from both sides to each other.how do you know when to do that move? and why didnt you instead ignore the 5 and 2 since its on both sides and write down the exponents alone as an entirely new equation? like (a-2)(3)=(2)(b)

[@GMATNinjaTutoring]

You should wait to do that move until you've one term with each base on each side of the equation, so only one 2 as a base on the left and right-hand side of the equation, only one 3, only one 5, etc. For example, I wouldn't make that move if I had (2^x)(3^y)(2^2) = (2^8)(3^7) as I've got two 2s as bases on the left-hand side of the equation. I'd want to take one more step to get to (2^[x + 2])(3^y) = (2^8)(3^7). You can then write down entirely new equations if you want to, but you should write one equation for each base. This means that in the example above, we'd have x + 2 = 8 for the terms with base 2, and we have y = 7 for the terms with base 3. What you can't do is combine the exponents from different bases into the same equation, so we couldn't have (x + 2)y = (8)(7). I hope that helps!

[@devamshidinesh4132]

@@GMATNinjaTutoring crystal clear ! thank you so much.

[@DhvanilChavda]

Hi,IN Q.6,WHAT IF WE DIVIDE 5^8X BY 5^4X then it will become 5^4x=100 and so answer can be 5^4x-2 = 4.Can you please tell me how this method i did was incorrect?

[@srilanka739]

question 4 was quite tough, is this level required to get a 44-45?

[@HalfGermany100]

Thank you so much!

[@IIBamboocha]

As always, thank you for the Video, I still have one question though, regarding Q4 (26:15): Ultimately, I would have to add the two terms 2^-8a + 2^-81. Since the base and the exponent are the same, I should be allowed to just add them, resulting in 4^-8a, which, in turn, I could factor out to (2^2)^-8a, which would result in 2^-16a. For sure, I made a mistake, but can you guys tell me where?

[@GMATNinjaTutoring]

Hi IIBamboocha, The mistake comes when you add the two terms and change the base. I'll give you two examples that will hopefully make the process a little more clear. If you add x^3 + x^3, we get 2 * x^3. The base, x, doesn't change, but we end up with two of what we had before. If we do this with numbers, for example adding 2^3 + 2^3, we get 8 + 8 = 16. This is the same as taking 2^3 + 2^3 and getting 2 * 2^3 = 2 * 8 = 16. We don't get 4^3 = 64. Using what we've seen in these examples, if we add 2^(-8a) + 2^(-8a), we get 2 * 2^(-8a). Now that we're multiplying two terms with the same base, we can add the exponents to get 2^(1-8a). I hoe that helps!

[@IIBamboocha]

@@GMATNinjaTutoring Damn, I see. Couldn't have asked for a better explanation. Thank you so much!

[@srilanka739]

is it possible to solve 5 by factoring out 5^a (1-1/25) = 5(24/25) 5(24/25)= 75 (2^b) what would the next step here be? :(

[@GMATNinjaTutoring]

Hi! We could actually do that, although it's probably not the most efficient path. From there, we could say that 5^a(24/25)=5^a(3*(2^3)/25)=(5^a)*3*(2^3)/(5^2)=3*(2^3)*(5^(a-2)). I then know that this is equal to 75*(2^b)=(5^2)*3*(2^b). So, 3*(2^3)*(5^(a-2))=(5^2)*3*(2^b). The 3s cancel. Because a and b are integers, I know that 5^2=5^(a-2), and a=4. I also know that 2^3=2^b, so b = 3. Therefore, a + b = 7. I hope that helps!

[@ggas33dfdf]

Q1: What happened to the "3" in 3^10, its gone... please explain.

[@GMATNinjaTutoring]

I think you're asking about the change as Bransen moved from the second to the third line of his solution, but please tell me if I've got that wrong and I can try to answer your question again. In the second line of his solution, Bransen had 3^(-6x) * 3^10. When you multiply two terms with the same base (in this case, 3), you can add the exponents. This means 3^(-6x) * 3^10 = 3^(-6x + 10). I hope that helps!

[@anuragmishra145]

2nd last has got me stumped

[@srilanka739]

just try test cases under the rules b^a>b>1 make b = 2 a=2 2^2>2>1 sub it back into the question stem you will always get a No answer - so A is sufficient as the answer is definitely NO statement 2 is very easy to dismiss

[@banafshehz6390]

But in Q4 couldn't we simply consider -4 as 2^2 and not (-2)^2 cause a is gonna be positive so why we go such a long road the first one is possitive 2 and for the same reason the sec one gonna be positive 2 and then factor out, easy 😁

[@GMATNinjaTutoring]

Yes, that's a good point! Since (-4) is raised to an even power, we know that it will end up positive. If you see that early on, that would definitely make things quicker. Thanks so much for the comment!

[@danielgomes1631]

Q6 Is wrong?? Bc 5 raised to anything cannot give 0 in the units place.... It will always be 5

[@GRENinjaTutoring]

Great question! You're correct that 5 raised to any power will have a 5 in the units place IF 5 is raised to an INTEGER. Notice that Q6 doesn't limit x to integer values. In fact, by using logarithms, you could solve for an exact value of x that would make 5^8x = 62,500. But it definitely wouldn't be an integer, and you don't need to do this to solve Q6. The good news is that you won't ever need to use logarithms on the GMAT. However, one common theme on the GMAT is limitations placed on the value of x. Sometimes x IS limited to integers. Sometimes, no limitations are placed on the value of x. Either way, those parameters are often key to evaluating a question. I hope that helps!

[@danielgomes1631]

@@GRENinjaTutoring Yes thanks a lot for your help

[@shriyajajula5478]

can someone elaborate on the Q8 as to how we got b>1 and a>1?

[@GRENinjaTutoring]

Statement 1 says that 11, then it would get bigger if it were raised to a power greater than 1. However, it would get smaller if raised to a power less than 1. Now, we also know that b^a>b. In other words, if we raise "b" to the "a" power, it gets bigger. Therefore, "a" must also be greater than one. I hope that helps!

[@rikhrajghosh9897]

Pls help me with Q.8 !!!!!

[@GMATNinjaTutoring]

Happy to help!! If you let me know where in the question you're having problems, I'll do all I can to explain things

[@rikhrajghosh9897]

57:43 from here I am not able to understand, pls help me

[@GMATNinjaTutoring]

Hi @@rikhrajghosh9897, at this point we've pushed the question to the point where we know its asking whether sqrt(a^b) > (a^b)^2 and we've made the substitution x = a^b to make the problem a little simpler for now. This means we're asking whether sqrt(x) > x^2 and when this might happen if we know that a and b (and therefore x) must be positive. We can say that sqrt(x) > x^2 when 0 < x < 1 and sqrt(x) > x^2 when 1 < x, so we can rephrase this question to ask whether 1 < x. At this point, we can undo the substitution, so the question we're asking is whether a^b < 1. We can go one step further because we know b is positive. If a > 1 then no matter what value b takes as long as it's positive, we know a^b > 1. Similarly, if a < 1 then no matter what value b takes as long as it's positive, we know a^b < 1. This means we can reduce the complexity of the question further, so we're now asking whether a < 1. Once we've done all that, we're ready to get into statements (1) and (2) but I'll leave things here for now. I hope all that makes sense and has helped a bit!

[@adki7559]

Hi Bransen, would you be so kind and elaborate what you did at kzbin.info/www/bejne/in3Ykml4j51sj7c? When I did the magic, I ended up at 5^(a-2) * 2^3 = 5^2 * 2^b and I was baffled how you just come to: hey from here it is easy because: a-2=2 so a=4 and b=3. WAIT WAIT WAIT. How can you do that? They have different bases, I expected it to be more like: trying to get the same base and then work with the exponents but I am still surprised that you just do it with different bases. Would be awesome if you could link an explanation or quickly summarize into a more general concept. BR ADKI

[@BresStephane]

I would love to know how he managed to get there too. I'm not able to find anyone online talking about that magic technique "/

[@GMATNinjaTutoring]

Hi Ad and Bres! Part of it is that I know that the numbers are integers, but part of it is algebra too. I can divide both sides by 5^2 * 2^b, and I get that 5^(a - 2) * 2^(3 - b) = 1. In theory, 5^(something) and 2^(something) could maybe somehow be a number and its reciprocal to multiply and be equal to one. But I know that a and b must be integers, and I also know that anything raised to 0 is equal to one. So, if a = 2 and b = 3, then the left side of the equation will be equal to one. I hope that helps!

[@adki7559]

@@GMATNinjaTutoring Hi again, thank you very much for clarifying in detail. I understand now, makes sense. Please correct me if I am wrong with this, but I think you made a typo in your calculations. I think it should be: 5^(a-2)*2^3 = 5^2 * 2^b --> divided by (5^2 * 2^b) 5^(a-4) * 2^(3-b) = 1 and now it totally makes sense that if we set a to 4 then 5^0=1 and the same with the other term 2^(3-3)=2^0=1 1*1=1. Thanks a lot! BR ADKI

[@BresStephane]

@@GMATNinjaTutoring Great explaination thanks !

[@neonlights4937]