Mind Blowing Math Olympiad Exponential Equation. 3^x - 2^x = 65
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- เผยแพร่เมื่อ 8 ก.พ. 2025
- The Math Olympiad is a competitive examination designed to challenge and inspire students with a strong interest and love for mathematics.
In this tutorial, you will #learnhowtosolvetheexponentialequation in #matholympaid Exams.
A good knowledge on the use of Exponential/Indcial Rules, Logarithms and solving of equations are added advantage for better solving of these exponential equations.
However, they were considered in this tutorial to encourage beginners who have zero skills and with different learning pace.
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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
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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Thank you for the link : I shall view the questions when I have time.
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.
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4
p=33 and q=32 would give p+q=65 and p-q=1. However, it is not a solution.
So, what is the problem with P=64.5 and Q = 0.5?
33+32=65, 33-32=1 : p=33 q=32
is another solution
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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.
Teach me African teacher ❤
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x=4
@@seydoukonate3868
Yes 👏
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.
@@edemafan4938
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X =4.
@@pas6295
Beautiful
3^x-2^x=65 3^x-2^x=8^2+1
3^4 -2^4=2^6+1 81-16=65
x=4
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65=81-16 and then from there blah blah you get x=4
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x=4
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X=4
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