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Bunuel
If x and y are integers and \(72^x∗54^y=96\), what is the value of x−y?

A. -1
B. 0
C. 1
D. 2
E. 3

72^x∗54^y=96 divide 6 at both side of the equation:
12^x*9^y=16 divide 3 at both side of the equation:

4^x*3^y=16/3
So x=2, y=-1

Hi Missyy

You cannot divide both sides of the exponents and use the quotients in a manner that you did.

for eg if we have \(6^2*9^2\) and as per you approach if I divide this by 3, then I should get \(2^2*3^2=36\)

but actually it will be \(\frac{6*6*9*9}{3}=972\).

Moreover the division is also wrong. \(72^x*54^y\) is one number but you are individually dividing both \(72^x\) & \(54^y\) by 6 individually. Had there been a + or - then individual division might be possible but in case of variable exponents it is incorrect
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Bunuel
If x and y are integers and \(72^x∗54^y=96\), what is the value of x−y?

A. -1
B. 0
C. 1
D. 2
E. 3

Be careful in such questions



\(72^x * 54^y = 96\)

Convert the above into their factors

[2^3 * 3^2]^x * [2*3^3]^y = 2^5 * 3^1
2^3x * 2^y * 3^2x * 3^3y = 2^5 * 3^1

3x + y = 5
2x + 3y = 1

Solve them to get x =2 & y =-1

E
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Bunuel
If x and y are integers and \(72^x∗54^y=96\), what is the value of x−y?

A. -1
B. 0
C. 1
D. 2
E. 3

My reasoning:

72 broken down into primes is \(2^3 * 3^2\) After multiplying x in we get \({2^3x} * 3^{2x}\)

54 broken down into primes is \(2 * 3^3\) After multiplying y in we get \(2^y * 3^{3y}\)

96 broken down into primes is \(2^5 * 3\)

Same base means you can add the exponents together so we get: \(2^{(3x+y)} * 3^{(2x+3y)}\)

We know \(2^{(3x+y)} = 2^5\) and \(3^{(2x+3y)} = 3^1\)

We can solve for x and y now which gives us 2 and -1 respectively. Subtract them and we get 3 (Choice E)
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My brain freezes with these kind of Q's. It was a guess answer but excluding the total, I did notice a constant play of '9's and '6's in the Q. So I felt the best answer could be 3. I may have gotten lucky.
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