A Very Nice Math Olympiad Question | How to solve 3^m-2^m=65 | Algebra

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See the way I breakdown the solution of this question. There is a lot you can learn from this video.
How to solve 3^m-2^m=65. ENJOY
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1 окт 2024

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Комментарии : 40

@lechaiku 4 месяца назад

If m E N there is a very fast solution: 3^m - 2^m = 65 Now we are looking for a power of 3 and power of 2 which their subtraction gives 65. 3^m - 2^m = 81 - 16 3^m - 2^m = 3^4 - 2^4 So m = 4 That's it.

@marcgriselhubert3915 4 месяца назад

Add that 3^m - 2^m is strictly increasing with m (m in N) and you can say that the solution is UNIQUE.

@user-u3e3kx5m6pg 4 месяца назад

First digit of 3ª is 3,9,7,1,3,9,7,1,•••• and 2ª is 2,4,8,6,2,4,8,6,••••. So first digit of 3ª-2ªis 5 if a=4n where n is a natural number. a=4 is a unique solution since 3ª-2ª is increasing.

@lechaiku 4 месяца назад

@@user-u3e3kx5m6pg Exactly, it's another way to find the solution.

@xgx899 4 месяца назад

@@marcgriselhubert3915 You need to prove the ``strictly increasing" part. One way to do it is through f'(x)=3^xlog 3-2^xlog 2. Another is by induction.

@marcgriselhubert3915 4 месяца назад

@@xgx899 It's evident that f'(x)>0 for x>0, because 3^x>2^x and ln(3)>ln(2), then you multiply these 2 inequations involving positive reals.

@kksomdr 2 месяца назад

In case2, if you use '-' not '+'. [ ]-[ ]=65-1. 2^m/2=2^5. Therefore m/2=5. m=10. But 3^10-2^10 is not 65.

@FrankSimonetta-rj4zf 4 месяца назад

This proof only works because m is even. This ensures that each of the factors evaluates to whole numbers, so they must be 5 and 13. If m were odd then m/2 would be a fraction which would make each of the factors irrational. Their product would still be 65 but you couldn't say one is 5 and the other is 13.

@ChrisBreemer 4 месяца назад

Cripes, a 12-minute video to "calculate" something you can just spot in a matter of seconds. This is so trivial that there's no way this was ever a Math Olympiad question.

@richardreiter6861 3 месяца назад

An easier way is to realize that 3^(m+1) -b^ (m+1) > 3 ^m-2^m There for only one possible answer Set 3^x - 2^ x = 81-16 = 65 X=4

@timk3539 4 месяца назад

A lot of comments are missing the point. It's irrelevant that this problem is easily solved by inspection. The solution presented is a general method.

@SpencersAcademy 4 месяца назад

👏💯

@barneynisbet6267 3 месяца назад

By inspection, m=4. Why is it a problem? For those suggesting a method is needed, m is defined as an integer. Therefore it’s easy to force a solution. When m is only real, or indeed complex, then a method is needed, and this isn’t it.

@jacquesperio3017 2 месяца назад

Par itération, en essayant 2, puis 3_ puis 4, on trouve que 4 est solution. Pas besoin de tarabiscotage!

@gnanadesikansenthilnathan6750 3 месяца назад

Got it with trial and error initially. Now its clear

@pasodirect 4 месяца назад

ahogy az okosok gondolják , és ahogy nekem tanították , let u=3^(m/2) , u^2=3^m , let v=2^(m/2) , v^2=2^m , 65=13*5 , u^2-v^2=65 , (u+v)(u-v)=13*5 , (u+v)+(u-v)=13+5 , 2u=18 , u=9 , 3^(m/2)=9 , 3^m=81 , m=ln81/ln3 , (4*ln3)/ln3=4 , solu. m=4 , test --> OK , 3^4-2^4=81-16 , 81-16= 65 , OK ,

@SpencersAcademy 4 месяца назад

Fantastic. I must say it's an amazing approach you've got here.

@pasodirect 4 месяца назад

@@SpencersAcademy I also learned on RU-vid, I like it too, thank you, hello!

@elmer6123 4 месяца назад

You really need to know this trick: m=ceiling(ln65/ln3)=4. Check: 3^4-2^4=81-16=65.

@bathurdeenmoulanamanardeen665 4 месяца назад

65=13×5=(9+4)(9-4) 3^m|2=9=3^2 M|2=2 M=4

@SpencersAcademy 4 месяца назад

Fantastic

@dougnettleton5326 4 месяца назад

I'm curious whether you actually looked any further because, even without the contraint, that m is a natural number, 3^(m/2) = 33 can't lead to a solution. 3^(m/2) = 33, taking the log of both sides leads to... (m/2)×ln(3) = ln(33) => m = 2×ln(33)/ln(3) => m > 6.365 3^6.365 - 2^6.365 > 1006, certainly not 65.

@SpencersAcademy 4 месяца назад

Yeah, I did. I am wondering as to how you got 33.

@dougnettleton5326 4 месяца назад

@SpencersAcademy I didn't get 33. You got it after dividing 66 by 2, and underneath after some discussion, you wrote "no solution." It's in your video at 12:02.

@АндрейЛюбавин-э4щ 4 месяца назад

@Burlongaming 4 месяца назад

which software is this thanks

@Burlongaming 4 месяца назад

which software is this thanks to

@iontecu8521 4 месяца назад

Sorry, 3 tom/2 +2tom/2

@kevingeier3385 4 месяца назад

M equals 3 which 3 power 3 whitish 81 and 2 power of 3 equals 16 Therefore 81 minus 15 equals 65

@AmanNgmOP 4 месяца назад

You mean 4?

@boogy2239 4 месяца назад

81-15=66

@nasrullahhusnan2289 4 месяца назад

Last digit of 3^m is 3,9,7,1 for m=1,2,3,4 and repeated for other m. Whle last digit of 2^m is 2,4,8,6 and repeated for other m. As (3^m)-(2^m)=65, meaning that the last digit of difference between the two is 5, it happens for m=2,4. For m=2, 3²-2²=5, does not satisfy the equation. For m=4, 3⁴=81 and 2⁴=16. So 3⁴-2⁴=81-16=65 as required. Thus m=4

@SpencersAcademy 4 месяца назад

That's an excellent delivery you've got. I like that

@iontecu8521 4 месяца назад

You know that 3to m+2to m is integar?

@iontecu8521 4 месяца назад

Sorry, 3 to m/2 + 2tom/2 is integer?

@nasrullahhusnan2289 4 месяца назад

@@iontecu8521: No mention that 3^m and 2^n are integers. Only when 3 and 2 are raised to integer m, which may not be integer. Suppose that m is not an integer. 3^m and 2^n are certainly not integers; they are irrational numbers. Note that their difference is an integer. Do you thimk that difference of two irrational numbers can be an integer? If the answer is 'NO' then the converse is true.