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26 Apr 2019, 13:18
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This question was posted as a DS question at one of the prep company with different numbers. You need to find x and y.
1. 4\(^\frac{x-y}{3}\)=\(\frac{1}{16}\)
2. product of xy is given
I marked A, thinking as follows:
\(3\sqrt{\frac{4^x}{4^y}}\)=\(\frac{1}{4^2}\) ->/\(^3\)
\(\frac{4^x}{4^y}\)=\(\frac{1^3}{4^2^*3}\)
Accordingly, I thought x=0 and y=6
Solution was C. Can someone please point out of to the flaw in my reasoning?
1. 4\(^\frac{x-y}{3}\)=\(\frac{1}{16}\)
2. product of xy is given
I marked A, thinking as follows:
\(3\sqrt{\frac{4^x}{4^y}}\)=\(\frac{1}{4^2}\) ->/\(^3\)
\(\frac{4^x}{4^y}\)=\(\frac{1^3}{4^2^*3}\)
Accordingly, I thought x=0 and y=6
Solution was C. Can someone please point out of to the flaw in my reasoning?
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26 Apr 2019, 13:31
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Your fundamentals are not clear. You have concept gaps and thereby need to revisit a few of the foundation concepts. Also, you need to provide the complete question when you are looking for a solution.
4\(^\frac{x-y}{3}\)=\(\frac{1}{16}\)
4\(^\frac{x-y}{3}\)=\(\frac{1}{4^{2}}\)
4\(^\frac{x-y}{3}\)=\(4^{-2}\)
\(\frac{x-y}{3}\) = -2
x - 3 = -6
Therefore, statement 1 provides infinite values of x and y. Not sufficient.
I am guessing statement 2 provide xy = constant. Therefore, statement 2 also provides infinite values of x and y. Not sufficient.
Statements 1 and 2 together provide unique values of x and y.
4\(^\frac{x-y}{3}\)=\(\frac{1}{16}\)
4\(^\frac{x-y}{3}\)=\(\frac{1}{4^{2}}\)
4\(^\frac{x-y}{3}\)=\(4^{-2}\)
\(\frac{x-y}{3}\) = -2
x - 3 = -6
Therefore, statement 1 provides infinite values of x and y. Not sufficient.
I am guessing statement 2 provide xy = constant. Therefore, statement 2 also provides infinite values of x and y. Not sufficient.
Statements 1 and 2 together provide unique values of x and y.
Thank you for your response, DisciplinedPrep
I understand your approach. But I don't understand why my approach is wrong. I think I did everything according to the exponents rules, maybe I can't solve this way because there is priority of order?
\(\frac{b^m}{b^n}\)=\(b^m^-^n\)
\(b^\frac{1}{m}\)=m\(\sqrt{b}\) (\(m^t^h\) root of b) -> ->
4\(^\frac{x-y}{3}\)=\(\frac{1}{16}\)
\(3\sqrt{\frac{4^x}{4^y}}\)=\(\frac{1}{4^2}\) ->/\(^3\)
\(\frac{4^x}{4^y}\)=\(\frac{1^3}{4^2^*3}\)
-> x=0 and y=6
Which of these steps is not correct? I revised my notes and I think my steps to reach \(\frac{4^x}{4^y}\)=\(\frac{1^3}{4^2^*3}\) are correct.
I think the problem might be that I can not conclude the solution (x=0, y=6) based on \(\frac{4^x}{4^y}\)=\(\frac{1^3}{4^2^*3}\) for some reason.
So I guess my real question is why cant I do so?
Thank you for your help!
R
*edit: I think I realized where I went wrong. I shouldn't have assume that 1=4^0. 4^0=1 but 1 itself cannot be assumed to be 4^0 in this case. Especially because the denominator is unknown. If the fraction would be 4^x/3=1/3 then x can only be 0. Since the denominator is unknown, I cannot assume x=0
I understand your approach. But I don't understand why my approach is wrong. I think I did everything according to the exponents rules, maybe I can't solve this way because there is priority of order?
\(\frac{b^m}{b^n}\)=\(b^m^-^n\)
\(b^\frac{1}{m}\)=m\(\sqrt{b}\) (\(m^t^h\) root of b) -> ->
4\(^\frac{x-y}{3}\)=\(\frac{1}{16}\)
\(3\sqrt{\frac{4^x}{4^y}}\)=\(\frac{1}{4^2}\) ->/\(^3\)
\(\frac{4^x}{4^y}\)=\(\frac{1^3}{4^2^*3}\)
-> x=0 and y=6
Which of these steps is not correct? I revised my notes and I think my steps to reach \(\frac{4^x}{4^y}\)=\(\frac{1^3}{4^2^*3}\) are correct.
I think the problem might be that I can not conclude the solution (x=0, y=6) based on \(\frac{4^x}{4^y}\)=\(\frac{1^3}{4^2^*3}\) for some reason.
So I guess my real question is why cant I do so?
Thank you for your help!
R
*edit: I think I realized where I went wrong. I shouldn't have assume that 1=4^0. 4^0=1 but 1 itself cannot be assumed to be 4^0 in this case. Especially because the denominator is unknown. If the fraction would be 4^x/3=1/3 then x can only be 0. Since the denominator is unknown, I cannot assume x=0
