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I'm a little rusty, how do you cross multiply to get everything to be a divisor of 1 to being the number alone to the exponent. That is where I struggled with this problem. Thanks.
I'm a little rusty, how do you cross multiply to get everything to be a divisor of 1 to being the number alone to the exponent. That is where I struggled with this problem. Thanks.
I rewrote the equation as: 7^-x∗8^-12=28^-18y 7^-x∗2^3(-12) = 7^-18y∗ 2^2(-18y) 7^-x∗2^-36 = 7^-18y∗ 2^-36y from here we can see that -x = -18y or x = 18y and -36=-36y or y =1. If y is 1 then x = 18 y-x = 1-18 = -17
Bunuel Hi Bunuel, if x & y are not integers, does it give infinitely many solutions of x & y, or does it still give finite (but different) solutions? Thanks!
Bunuel Hi Bunuel, if x & y are not integers, does it give infinitely many solutions of x & y, or does it still give finite (but different) solutions? Thanks!
It will have infinite number of solutions. For every x, there will be some y to satisfy the equation and vise-versa.
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i did this question in following way: simplified the given equation into 7^(18y-x) = 2^36(1-y) now as 7 and 2 are prime no.s , LHS will be equal to RHS only when when respective powers of 7 and 2 is 0 hence, i put 18y-x=0, this implies 18y=x. and 36(1-y)=0, this implies y=1 hence y-x= 1-18= -17
i did this question in following way: simplified the given equation into 7^(18y-x) = 2^36(1-y) now as 7 and 2 are prime no.s , LHS will be equal to RHS only when when respective powers of 7 and 2 is 0 hence, i put 18y-x=0, this implies 18y=x. and 36(1-y)=0, this implies y=1 hence y-x= 1-18= -17
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61% (01:34) correct 
