A note on nomenclature: a power is a base raised to an exponent. For example: 8 is the power of base 2 with exponent 3. For example: When you multiply powers with the same base you add exponents. In the video the word power is used in both senses- as power and as exponent. The exponents are sometimes called also the indices. Words are important, it is easy to sow confusion with wrong wording.
On the problem 3^2x -2^2y= 77 An easier solution is found if you change 77 to 81-4 and solve. x=2 and y=1 Does either method prove there are no other answers? Is there any other. 3^n - 2 ^m = a-b where a-b=77 where n &m are integers
Well brother , if we write 81-4 instead of 77 and make it as 3^2x - 2^2y = 77 3^2x - 2^2y = 81-4 3^2x - 2^2y = 3^4 - 2 ^2 3^2(2) - 2^2(1) = 3^4 - 2^2 81-4 = 81-4 77 = 77 Then x = 2 and y = 1 Whether it is applicable mathematically , at first right after seeing the problem i thought like this , and got x as 2 and y as 1 , just to make sure whether my answers were right i entered into the video and confirmed that my answers were right , but now i am in a confusion whether my steps were right , so can you tell me whether the way i solved this problem is right or wrong
Okay, the solution you propose is correct but, for me, less elegant. Mathematics is also something that has to be elegant. So I prefer the solution in the video.
No need of pen and paper can be solved directly , like since x and y are positive integers x can be either 1 or 2 it cannot be 1 as something's is subtracted from it hence for sure x is 2 now putting x as 2 we get y as 1 hence x=2 y=1
Here is how I get my answer in my previous comment: [3^(2x)]-[2(2^y)]=77 --> (3^m)-(2^n)=77 where m=2x and n=2y Last digit of 3^m is (3,9,7,1) for m=(1,2,3,4) and repeated for other m, while last digit of 2^n is (2,4,8,6) for n=(1,2,3,4) and repeated for futher n. As (3^m)-(2^n)=77, meaning its last digit is 7, then last digit of 3^m must be 1 (m=4) and last digit of 2^n must be 4 (n=2). Thus 4=m=2x --> x=2 and 2=n=2y --> y=1
X=2 Y= 1
1. X1 = 2, Y1 = 1
2. X2 = log3 of 39, Y2 = log2 of 38
Which program are you using to write?
A note on nomenclature: a power is a base raised to an exponent. For example: 8 is the power of base 2 with exponent 3. For example: When you multiply powers with the same base you add exponents. In the video the word power is used in both senses- as power and as exponent. The exponents are sometimes called also the indices. Words are important, it is easy to sow confusion with wrong wording.
On the problem 3^2x -2^2y= 77
An easier solution is found if you change 77 to 81-4 and solve. x=2 and y=1
Does either method prove there are no other answers?
Is there any other. 3^n - 2 ^m = a-b where a-b=77 where n &m are integers
Entendi tudo usando os olhos e tapando os ouvidos. Ou seja : deixando o celular sem som.
Why 2 to the power y - (- 2 to the power y) is not equal 4 to the power y?
Well brother , if we write 81-4 instead of 77 and make it as
3^2x - 2^2y = 77
3^2x - 2^2y = 81-4
3^2x - 2^2y = 3^4 - 2 ^2
3^2(2) - 2^2(1) = 3^4 - 2^2
81-4 = 81-4
77 = 77
Then x = 2 and y = 1
Whether it is applicable mathematically , at first right after seeing the problem i thought like this , and got x as 2 and y as 1 , just to make sure whether my answers were right i entered into the video and confirmed that my answers were right , but now i am in a confusion whether my steps were right , so can you tell me whether the way i solved this problem is right or wrong
Weldone, brother. Your steps are absolutely correct. In fact, most mathematicians do use this trick in solving some complex maths problems.
@@SpencersAcademy well then why don't you use this thing as an alternative solution in the end of your video , so that it would be way more easier
@@nalinivijayan5617 That's a good idea
@@SpencersAcademy Thank you , hope you do that next time brother
Okay, the solution you propose is correct but, for me, less elegant. Mathematics is also something that has to be elegant. So I prefer the solution in the video.
I am your 1k th subs😊
I am so grateful, man. My heart is doing the happy dance. 💃 💃 💃 💃 💃 💃 💃
x=2, y=1. x= -1 y=-2 etc four pairs of solutions . Now I'll watch and see if I am missing something
Yes, i was too hasty: .x=-2, y=-1 x=-2, y=+1 x=+2, y=-1, and x=+2, y=+1
No I was guessing right in the first ten seconds. I think I am getting too old for all the excitement of new tricks. Lot less than 14 minutes here.
No need of pen and paper can be solved directly , like since x and y are positive integers x can be either 1 or 2 it cannot be 1 as something's is subtracted from it hence for sure x is 2 now putting x as 2 we get y as 1 hence
x=2
y=1
X n Y = 2
(1+log(3)13;1+log(2)19)(2;1)
Nice one
Lindo!!!!!
Thanks, bro. I'm glad you enjoyed it.
x=2 and y=1
Excellent
Here is how I get my answer in my previous comment:
[3^(2x)]-[2(2^y)]=77 --> (3^m)-(2^n)=77 where m=2x and n=2y
Last digit of 3^m is (3,9,7,1) for m=(1,2,3,4) and repeated for other m, while last digit of 2^n is (2,4,8,6) for n=(1,2,3,4) and repeated for futher n.
As (3^m)-(2^n)=77, meaning its last digit is 7, then last digit of 3^m must be 1 (m=4) and last digit of 2^n must be 4 (n=2). Thus
4=m=2x --> x=2 and 2=n=2y --> y=1