Bunuel
If \(\frac{10^{2x}}{100^{3y}} < 1\), which of the following must be true?
A x < 3y
B 2x < 3y
C 2x < y
D 2x + 3y < 1
E 2x + 3y < 0
To determine which statement must be true, we need to simplify the given inequality using exponent rules.
Given inequality:
\(\frac{10^{2x}}{100^{3y}} < 1\)
Method 1: Algebraic Approach (Base Conversion)The key is to express both the numerator and the denominator with the same base.
We know that \(100 = 10^2\).
Substitute this into the denominator:
\(\frac{10^{2x}}{(10^2)^{3y}} < 1\)
Apply the power rule \((a^b)^c = a^{b \cdot c}\):
\(\frac{10^{2x}}{10^{6y}} < 1\)
Apply the quotient rule \(\frac{a^m}{a^n} = a^{m-n}\):
\(10^{2x - 6y} < 1\)
We know that \(1\) can be written as \(10^0\).
\(10^{2x - 6y} < 10^0\)
CRITICAL RULE: Since the base (10) is greater than 1, we can drop the bases and compare the exponents directly without flipping the inequality sign.
\(2x - 6y < 0\)
Add \(6y\) to both sides:
\(2x < 6y\)
Divide by 2:
\(x < 3y\)
This matches Option (A).
***
Method 2: Picking NumbersLet's find values for \(x\) and \(y\) that make the original inequality true, then test the options to eliminate incorrect ones.
We need \(\frac{10^{2x}}{100^{3y}} < 1\).
Let's try \(y = 1\).
The denominator becomes \(100^3 = (10^2)^3 = 10^6\).
To make the fraction less than 1, the numerator \(10^{2x}\) must be smaller than \(10^6\).
So, let's choose \(2x = 4\), which means \(x = 2\).
Check validity: \(\frac{10^4}{10^6} = 10^{-2} < 1\). (Valid set: \(x=2, y=1\)).
Now test the options:
A. \(x < 3y \implies 2 < 3(1)\) -> \(2 < 3\) (
True)
B. \(2x < 3y \implies 2(2) < 3(1)\) -> \(4 < 3\) (
False)
C. \(2x < y \implies 4 < 1\) (
False)
D. \(2x + 3y < 1 \implies 4 + 3 < 1\) -> \(7 < 1\) (
False)
E. \(2x + 3y < 0 \implies 7 < 0\) (
False)
Only Option A remains.
Answer: A