+2653 Find the minimum value of 9^x - 2 \cdot 3^x + 1 over all real numbers x.
+16 \(9^{x}-2\cdot 3^{x}+1\):
Set a = \(3^x\)
\(a^2-2a+1=0\)
\((a-1)^2=0 \)
a=1, so \(3^x\)=1
That means x=0 because anything to the 0th power is one
Hope this helps!
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