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.
For integers x and y, 25^(−10)*100^x=2^y. What is the valu
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Replace 25 and 100 with their prime factors. 25 = 5^2 and 100 = 5^2 x 2^2vinni.k wrote: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.
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
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25¯¹� * 100^x = 2^yvinni.k wrote: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
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.
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25^(-10) * 100^x = 2^yvinni.k wrote: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) * 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
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