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

I am your 1k th subs😊

@SpencersAcademy

Ай бұрын

I am so grateful, man. My heart is doing the happy dance. 💃 💃 💃 💃 💃 💃 💃

Entendi tudo usando os olhos e tapando os ouvidos. Ou seja : deixando o celular sem som.

X n Y = 2

1. X1 = 2, Y1 = 1 2. X2 = log3 of 39, Y2 = log2 of 38

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.

@richardreiter6861

5 күн бұрын

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

Lindo!!!!!

@SpencersAcademy

Ай бұрын

Thanks, bro. I'm glad you enjoyed it.

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

@SpencersAcademy

Ай бұрын

Weldone, brother. Your steps are absolutely correct. In fact, most mathematicians do use this trick in solving some complex maths problems.

@nalinivijayan5617

Ай бұрын

@@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

@SpencersAcademy

Ай бұрын

@@nalinivijayan5617 That's a good idea

@nalinivijayan5617

Ай бұрын

@@SpencersAcademy Thank you , hope you do that next time brother

@BruceLee-io9by

Ай бұрын

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.

(1+log(3)13;1+log(2)19)(2;1)

@SpencersAcademy

Ай бұрын

Nice one

x=2 and y=1

@SpencersAcademy

Ай бұрын

Excellent

@nasrullahhusnan2289

Ай бұрын

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