\(3^x+1=9^x\)
↔\(3^{2x}-3^x=1\)
↔\(3^x\left(3^x-1\right)=1\)
↔\(3^x;3^x-1\in\left\{-1;1\right\}\)
Ta có bảng sau:
\(3^x\) | -1 | 1 | |
\(3^x-1\) | -1 | 1 | |
x | x\(\in\varnothing\) | x\(\in\varnothing\) |
\(3^{x+1}=9^x\)
\(3^{x+1}=\left(3^2\right)^x\)
\(3^{x+1}=3^{2x}\)
\(\Rightarrow x+1=2x\)
\(2x-x=1\)
\(x=1\)
Vậy \(x=1\)
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