2^x*3^5 = 2^(2y−1)*3^(2y+1)

Question

2^x*3^5 = 2^{(2y−1)}*3^{(2y+1)}

If x and y are positive integers, then x =

A. 2
B. 3
C. 4
D. 5
E. 6

Anisha1637 · Answer

We can equate the exponents as given below:
x=2y-1------(i)
5=2y+1------(ii)
Solving (ii) we get y= 2
5-1=2y
2=y

Putting the value of y in (i)
We get x=2*2-1
x=3
Therefore answer is B

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Kinshook · Answer

Given:  2^x*3^5 = 2^{(2y−1)}*3^{(2y+1)}

Asked: If x and y are positive integers, then x =

Equating powers of 2
2y - 1 = x
2y = x+1

Equating powers of 3
5 = 2y + 1
5 = (x+1) + 1
x = 3

IMO B

Nidzo · Answer

2^x ∗ 3^5= 2^(2y−1)∗ 3^(2y+1)

3^5 = 3^(2y+1)
y =2

Plugging y=2 into 2^x = 2^(2y−1) gives that x = 3

Answer B

MathRevolution · Answer

2y + 1 = 5

2y will be equal to 4

2y-1 will be 3

x = 3

Answer B

2^x3^5 = 2^(2y−1)3^(2y+1)