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\[3^{2x} -12 \times 3^x +27 =0 \]
I got \[3^{2x}-12 \times 3^x + 3^3 =0\] But I don't know what I can do with the -12
substitute u = 3^x \[u^2 - 12u +27 = 0\]
\[(u-9)(u-3)=0\]
\[u=3,9\]\[3^1=3^x or 3^2 = 3^x\] x = 1, 2
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you got it.... correct
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