How to solve a Nice Math Olympiad exponential problem. The laws of indices were applied while solving this problem. #matholympiadproblem #mathcompetition #education #exponential #equations
Пікірлер: 25
@suddhasheelАй бұрын
You could have directly assumed the odd / even logic right at the beginning. Since the answer is odd, 8^(y-8) must be 8^(0). Thus directly y = 8. Then apply the remaining logic to derive the value of x.
@johnmaulkin-ur8lmАй бұрын
Excellent , again well presented. Thank you.
@JJONLINEMATHSCLASSchannelАй бұрын
So nice of you
@clloydmathison995Ай бұрын
You have a very pleasant personality for teaching. Kudos!!!❤❤❤❤❤❤❤❤❤❤
@JJONLINEMATHSCLASSchannelАй бұрын
Thank you! 😃
@umarbashir1236Ай бұрын
Thank you,
@dumitrudraghia5289Ай бұрын
OBS. 8^(a+y-8)-8^(y-8)= 4095; 8^(y-8)x(8^ a-1)=4095. Etc.
@devonwilson5776Ай бұрын
Greetings. Great job. Absolutely brilliant.
@JJONLINEMATHSCLASSchannelАй бұрын
Thank you! Cheers!
@bartconnolly6104Ай бұрын
Fairly simple to do in your head . Most people know 2^10 =1024 so,2^12 =4096 which is 4095 +1 2^12 =2^3^4=8^4 and 1= anything ^0 including 8^0 => 8^4-8^0 =4095 and the rest is easy The driving factor wa"what is the lowesrpower of 2 which is greater than 4095 but divisible by 3 ?' Divisible by three means it is an power of 8,
@user-sr7uc1xo3iАй бұрын
Μπράβο σου. Είσαι καταπληκτική. Από Ελλάδα. Greece
@JJONLINEMATHSCLASSchannelАй бұрын
Thanks
@mouradbelkas598Ай бұрын
Good job. Thank you
@JJONLINEMATHSCLASSchannelАй бұрын
Our pleasure!
@olayinkaadeleye3134Ай бұрын
Elegant
@JJONLINEMATHSCLASSchannelАй бұрын
Thanks
@sylvesterogbolu-otutu1498Ай бұрын
The solution set is x, y = {12, 8} Working Method. My approach would be to rewrite the right side with a base of 8 - since 4095 = 4096 - 1. 4096 = 2^12 or 8^4; and 1 = 8^0. Therefore; 4095 = 8^4 - 8^0 Accordingly; The entire expression can be presented as: 8^(x - 8) - 8^(y - 8) = 8^4 - 8^0 From this, we can equate exponents and solve the resulting linear equations, thus: x - 8 = 4, and y - 8 = 0 If x - 8 = 4, x = 4 + 8; x = 12 If y - 8 = 0, y = 8 + 0; y = 8 Therefore, the final solution set is x, y = {12, 8} Checking: x = 12, and y = 8 8^(12 - 8) - 8^(8 - 8) = 4095 8^4 - 8^0 = 4095 4096 - 1 = 4095 (Checked).
@prollysineАй бұрын
A unique solution in my style... , let u=8^((x-8)/2) , v=8^((y-8)/2) , u^2=8^(x-8) , v^2=8^(y-8) , u^2-v^2=4095 , (u+v)(u-v)=4095 , etc. , 4095=105*39 , let u+v=105 , u-v=39 , u+v+u-v=105+39 , 105+39=144 , -> 2u=144 , u=72 , u+v=105 , v=105-72 , v=33, u^2=72^2 , u^2=5184 , v^2=33^2 , v^2=1089 , for x , -> 8^(x-8)=5184 , (x-8)*ln8=ln5184 , x-8=ln5184/ln8 , x-8=4.11328 , x=4.11328+8 , x=12.11328 , for y , -> 8^(y-8)=1089 , (y-8)=ln1089/ln8 , y-8=3.36293 , y=11.36293 , test , 8^(x-8)-8^(y-8)=8^(12.11328-8)-8^(11.3693-8) , 8^(12.11328-8)-8^(11.3693-8)=8^(4.11328) - 8^(3.36293) , test -> , 8^4.11328=5184 , 8^3.36293=1089 , 5184-1089=4095 , same , OK , solu. , x=12.11328 , y=11.36293 , /// Large numbers can be tamed by using logarithms... /// , By dividing the problem into products of 4095, there are many, many solutions... etc. , 4095=5*3*7*3*13 , -> 15*273=4095,,, ,