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Bunuel
If 10^x equals 0.1 percent of 10^y, where x and y are integers, which of the following must be true?

A. y = x + 2
B. y = x + 3
C. x = y + 3
D. y = 1,000x
E. x = 1,000y

\(10^x\) \(=\) \(\frac{0.1}{100}\) \(*\) \(10^y\)

\(10^x = \frac{1}{1000} * 10^y\)

\(10^x = 10^{-3} * 10^y\)

\(x = -3 + y\)

\(y = x + 3\)

(B)
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selim
Bunuel
If 10^x equals 0.1 percent of 10^y, where x and y are integers, which of the following must be true?

A. y = x + 2
B. y = x + 3
C. x = y + 3
D. y = 1,000x
E. x = 1,000y


Given

10^x equals 0.1 percent of 10^y.

So, we can translate into equation.


\(10^x\) = 0.1% * \(10^y\)

\(10^x\) = (0.1 /100)* \(10^y\)

\(10^x\) = 0.001 *\(10^y\)

\(10^x\) = 1*\(10^{y-3}\)

\(10^x\) = \(10^{y-3}\)

x = y-3

y = x+3

Thus the best answer is B.

Thank you for the explanation, that made the problem quite clear.

Although, I have to admit it feels weird learning math from Johnny Sins. :lol:
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Bunuel
If 10^x equals 0.1 percent of 10^y, where x and y are integers, which of the following must be true?

A. y = x + 2
B. y = x + 3
C. x = y + 3
D. y = 1,000x
E. x = 1,000y

10^x=10^-3*10^y
x=-3+y
y=x+3
IMO B
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there is another way to solve this problem. Although more difficult, it came out of my mind when i first saw this question.

When approaching this problem using logarithms, we can use the properties of logarithms to simplify the equation. First, we start with the given equation:
10^x=0.001×10^y
We can take the logarithm of both sides with base 10 (common logarithm):
x=log10(0.001×10^y)
Using the property of logarithms log(ab)=loga+logb, we get:
x=log10(0.001)+log10(10^y)
so x=-3+y
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