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Title: Solving logarithmic equations Post by: Mathguy3000 on Jul 24, 2018 2(27)^x = 9^x+1
Title: Re: Solving logarithmic equations Post by: bio_man on Aug 2, 2018 \(2(27)^x=9^x+1\)
\(2(3\cdot 9)^x=9^x+1\) \(2(3^x\cdot 9^x)=9^x+1\) \(2(3^x\cdot \left(3\cdot 3\right)^x)=9^x+1\) \(2(3^x\cdot 3^x\cdot 3^x)=3^x\cdot3^x+1\) 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, \(x = 1\). |