For real x, 3^x - 2^x is positive only for x > 0. For x>0 , both 2^x and 3^x are monotonically increasing functions, with 3^x increasing more rapidly. Thus, if any solution exists, it is unique. Here, by inspection, since 65 = 81-16, we see that x=4.
Toujours définir l'ensemble dans lequel le problème est posé. Etablir que le membre gauche de l'égalité est toujours croissant. Trouver une solution qui sera unique.
Thank you so much. I'm not even a student and i enjoyed learning from you. I have a question though. Why did you cross multiply at the end? I don't understand the logic behind this. If two numbers are equal to each other, why do we need to multiply them? What's the name of this method so I can read about it?
Good explanation but your method has a lack of demonstration at the begin. I think it comes from the fact we don't deal with this kind if problem in our academic studies (olympiad competitions). Before writting (3^(x/2))^2 -(2^(x/2))^2 =65, you must show that x is even by writting modulo 5 : 3 = -2(mod 5) 3^x-2^x = -2^(x+1) (mod5) so 3^x-2^x = 5k+1 or 5k-1 if x is odd 3^x-2^x = -2^x + 2^(x) (mod5) So 5 | 3^x-2^x. also p-q=1 and p+q= 65 p=33 and q=32. This is not solution because 33 is not a power of 3 even if 32 is a power of 2. Good lock.
For further studies on #matholympiad questions, kindly click the link below: th-cam.com/play/PLVcWSTbc_4a2v3prPWnDhJSi--js_DBbw.html&si=-vd_3gY2MlPcAUQ7
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65= 81- 16 chosen brcause 81 is a power of 3 and 16 is a power of 2 . Happily , 2^4 is 16 and 3^4 is 81 . Than x is 4 .
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Danger
If 3^x-2^x=65
Then 3^x>65
Then x>=4,
If x=4
Then 3^4=81 and 2^4=16,
81-16=65
X=4
That's exactly how I did this.
@@wedeyhabrom
This works if its objective question where the correct opinion is required ok. If it’s Theory, I’m afraid, you may not have good marks
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Solutions to such equations give √-1 which is i. This soln is so beautiful that it appears to be contrived!
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33+32=65, 33-32=1 : p=33 q=32
is another solution
Exactly what I want to say. Well-done 👍
For real x, 3^x - 2^x is positive only for x > 0. For x>0 , both 2^x and 3^x are monotonically increasing functions, with 3^x increasing more rapidly. Thus, if any solution exists, it is unique. Here, by inspection, since 65 = 81-16, we see that x=4.
@@RashmiRay-c1y
Great! But, in Theory , all works must be shown
Hello beautiful people and welcome back to our today’s class
Toujours définir l'ensemble dans lequel le problème est posé.
Etablir que le membre gauche de l'égalité est toujours croissant.
Trouver une solution qui sera unique.
You are doing a great job by explaining every step... Which Algebra book would you recommend to learn all the fundamentals for Math Olympics?
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Thank you so much. I'm not even a student and i enjoyed learning from you. I have a question though. Why did you cross multiply at the end? I don't understand the logic behind this. If two numbers are equal to each other, why do we need to multiply them? What's the name of this method so I can read about it?
4
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So, what is the problem with P=64.5 and Q = 0.5?
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x=4
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p=33 and q=32 would give p+q=65 and p-q=1. However, it is not a solution.
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X =4.
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Good explanation but your method has a lack of demonstration at the begin. I think it comes from the fact we don't deal with this kind if problem in our academic studies (olympiad competitions). Before writting (3^(x/2))^2 -(2^(x/2))^2 =65, you must show that x is even by writting modulo 5 :
3 = -2(mod 5)
3^x-2^x = -2^(x+1) (mod5) so 3^x-2^x = 5k+1 or 5k-1 if x is odd
3^x-2^x = -2^x + 2^(x) (mod5)
So 5 | 3^x-2^x.
also p-q=1 and p+q= 65 p=33 and q=32. This is not solution because 33 is not a power of 3 even if 32 is a power of 2.
Good lock.
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3^x-2^x=65 3^x-2^x=8^2+1
3^4 -2^4=2^6+1 81-16=65
x=4
@@anestismoutafidis4575
Your answer is correct , but you know that, you didn’t your methods.
In Mathematics, it is not acceptable okay
65=81-16 and then from there blah blah you get x=4
@@YolymaticsTutorials-kl6sx
Thanks a lot
For further studies on #matholympiad questions, kindly click the link below:
th-cam.com/play/PLVcWSTbc_4a2v3prPWnDhJSi--js_DBbw.html&si=-vd_3gY2MlPcAUQ7
Thank you for the link : I shall view the questions when I have time.
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@@belayabera2448
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x=4
@@ChristmasMaheri
Correct 👍 🍾🍾🍾
X=4