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question 2
for real numbers x and y, (4^x 4^y = 8^{xy})
○ true
○ false
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Explanation:

Step1: Simplify left - hand side

Using the rule of exponents \(a^{m}\cdot a^{n}=a^{m + n}\), for \(4^{x}\cdot4^{y}\), we have \(4^{x}\cdot4^{y}=4^{x + y}\). And since \(4 = 2^{2}\), then \(4^{x + y}=(2^{2})^{x + y}\). Using the rule \((a^{m})^{n}=a^{mn}\), we get \((2^{2})^{x + y}=2^{2(x + y)}=2^{2x+2y}\).

Step2: Simplify right - hand side

Since \(8 = 2^{3}\), for \(8^{xy}\), we have \(8^{xy}=(2^{3})^{xy}\). Using the rule \((a^{m})^{n}=a^{mn}\), we get \((2^{3})^{xy}=2^{3xy}\).

Step3: Compare the two sides

We have the left - hand side as \(2^{2x + 2y}\) and the right - hand side as \(2^{3xy}\). For the equation \(4^{x}4^{y}=8^{xy}\) to hold, we need \(2x + 2y=3xy\) for all real numbers \(x\) and \(y\). Let's test with \(x = 1\) and \(y = 1\). Left - hand side exponent: \(2(1)+2(1)=4\), right - hand side exponent: \(3(1)(1)=3\). Since \(2^{4}
eq2^{3}\) (because \(16
eq8\)), the equation does not hold for all real numbers \(x\) and \(y\).

Answer:

False