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Mathguy3000 Mathguy3000
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5 years ago
2(27)^x = 9^x+1
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Educator
5 years ago
\(2(27)^x=9^x+1\)

\(2(3\cdot 9)^x=9^x+1\)

\(2(3^x\cdot 9^x)=9^x+1\)

\(2(3^x\cdot \left(3\cdot 3\right)^x)=9^x+1\)

\(2(3^x\cdot 3^x\cdot 3^x)=3^x\cdot3^x+1\)

Set \(u=3^x\):

\(2(u^3)=u^2+1\)

\(2u^3-u^2-1=0\)

Something's wrong...

Let's try again...

\(2(27)^x=9^x+1\)

\(2(3^3)^x=9^x+1\)

\(2(3^x)^3=\left(3\cdot 3\right)^x+1\)

\(2(3^x)^3=3^x\cdot 3^x+1\)

Set \(u=3^x\):

\(2(u)^3=u^2+1\)

Yikes, we've got a cubic function! You could graph it...

That's your answer, \(x = 1\).
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