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question 10
for positive real number ( x ), ( 3^{-x} = 3^{1/x} ).
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false
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Explanation:

Step1: Analyze the exponents

For exponential functions with the same base, \(a^m = a^n\) implies \(m = n\) (when \(a>0,a
eq1\)). Here, the base is \(3\) (which is \(>0\) and \(
eq1\)). So we check if \(-x=\frac{1}{x}\) for positive real \(x\).

Step2: Solve \(-x=\frac{1}{x}\) for positive \(x\)

Multiply both sides by \(x\) (since \(x>0\), we can do this without changing inequality direction): \(-x^2 = 1\), or \(x^2=- 1\). But for real positive \(x\), \(x^2>0\), so \(x^2=-1\) has no real solutions, let alone positive real solutions. So \(3^{-x}
eq3^{\frac{1}{x}}\) for positive real \(x\).

Answer:

False