For integers x and y, 25^(−10)*100^x=2^y. What is the valu

[vinni.k]

For integers x and y, 25^(−10) * 100^x = 2^y. What is the value of y?

A. -10
B. 10
C. 20
D. 25
E. 40

OA is C

Please let me know how to approach. I am getting stuck in the middle.

[regor60]

For integers x and y, 25^(−10) * 100^x = 2^y. What is the value of y?

A. -10
B. 10
C. 20
D. 25
E. 40

OA is C

Please let me know how to approach. I am getting stuck in the middle.

Replace 25 and 100 with their prime factors. 25 = 5^2 and 100 = 5^2 x 2^2

So 25^-10 = 5^-20 and 100^x = 5^2x x 2^2x.

So the revised statement now reads: 5^(-20) x 5^(2x) x 2^(2x) = 2^y

Collect the powers of 5 and restate: 5^(2x-20) x 2^(2x) = 2^y

Since the right side doesn't contain any powers of 5, that must mean 5^(2x-20) equals 1. The way this equals 1 is for the exponent to equal 0.

2x-20 = 0 therefore X=10

Substitute X=10 back in

5^(20-20) x 2^(20) = 2^y

= 1 x 2^20 = 2^y

therefore Y= C, 20

[GMATGuruNY]

For integers x and y, 25^(−10) * 100^x = 2^y. What is the value of y?

A. -10
B. 10
C. 20
D. 25
E. 40

25¯¹� * 100^x = 2^y

1/25¹� * (25*4)^x= 2^y

(1/25¹�)(25^x)(4^x) = 2^y

[(25^x)/25¹�] * 4^x = 2^y

In the resulting blue equation, x and y will be integers if x=10, with the result that (25^x)/25¹� = 1.
Substituting x=10 into the blue equation, we get:

(25¹�/25¹�) * 4¹� = 2^y

4¹� = 2^y

(2²)¹� = 2^y

2²� = 2^y

Since each side of the red equation has the same base, the exponents must be EQUAL, with the result that y=20.

The correct answer is C.

[vinni.k]

Thanks

[Jeff@TargetTestPrep]

For integers x and y, 25^(−10) * 100^x = 2^y. What is the value of y?

A. -10
B. 10
C. 20
D. 25
E. 40

25^(-10) * 100^x = 2^y

25^(-10) * 25^x * 4^x = 2^y

Since the powers of 25 cancel out, x must be 10. Thus, we have:

4^10 = 2^y

(2^2)^10 = 2^y

2^20 = 2^y

20 = y

Answer: C