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If \(x\) and \(y\) are positive integers and \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\), then what is the value of \(y-x\)?
A. -18 B. -17 --> correct: \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\) => \(7^x*8^{12}={28}^{18y}\)=> \(7^x*2^{36}=7^{18y}*2^{36y}\) => x=18y & 36=36y => y=1 & x=18, so y-x=1-18=-17 C. 1 D. 17 E. 18
When trying to equate the powers of 2 on both sides to find Y, how did you get rid of 7^18 and 7^x?
Factorize: 2^36y∗7^18y=7^x∗2^36;
Equate the powers of 2 on both sides: 36y=36 --> y=1y=1;
Equate the powers of 7 on both sides: 1=x --> x=18x=18;
We did not get rid of anything there. Since given that x and y are positive integers, then in order \(2^{36y}*7^{18y}=7^x*2^{36}\) to be true, the powers of 2 should be equal and the powers of 7 should be equal. _________________
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