F.3 Maths laws of indices @ oob73zj92b的部落格

標題:

F.3 Maths laws of indices

發問:

(2^x)(6^y)=32(3^x+1) find the value of x

最佳解答:

I guess your 3^x+1 means 3^(x+1) 6^y = (2^y)(3^y) 32 = 2^5 hence 2^(x+y)3^y = 2^5(3^(x+1)) since prime factorization is unique we have x+y = 5 and y=x+1 so x = 2 , y=3 If your 3^x+1 really means 3^x + 1 then I can show you that it is impossible 2^(x+y)3^y = 2^5(3^x +1) observe that (3^x + 1) is not divisible by 3 so we must have 3^y = 1 hence y = 0 now LHS is 2^x , RHS is 2^5(3^x + 1) 2^(x-5) = 3^x + 1 RHS is an integer so x >= 5 now try x = 6 , it is clearly wrong. So, x must be greater than 6 Then 4 must divide LHS and hence 4 divide 3^x + 1 however, 3^x + 1 is divisible by 4 if and only if x is an odd number so x cannot be even, finally if x is an odd number then we have 3^x + 1 = (3+1)(1- 3 + 3^2 - 3^3 + ... + 3^(x-1) ) This term must be an odd number (1- 3 + 3^2 - 3^3 + ... + 3^(x-1) ) since there are total x odd terms, the sum of them is odd now we have 2^(x-5) = (3+1)(1- 3 + 3^2 - 3^3 + ... + 3^(x-1) ) which means 2^(x-5) is divisible by an odd number, impossible So there is no such x and y