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solve algebraically for x: 27^x=9^x+2
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we can rewrite this equation to (3^3)^x=(3^2)^x + 2
\[\left( 3^{3} \right)^{x}=\left( 3^{2} \right)x+2\] \[3^{3x}=3^{2x}+2\] \[\left( 3^{x} \right)^{3}=\left( 3^{x} \right)^{2}+2\] substitute \[3^{x}=y\] \[y ^{3}=y ^{2}+2\] \[y ^{3}-y ^{2}-2=0\] now you can solve for y and then for x
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