Solve the equation: 9^x-3^(x+1)+1=0

Question
Complex numbers
asked 2020-11-06
Solve the equation:
\(\displaystyle{9}^{{x}}-{3}^{{{x}+{1}}}+{1}={0}\)

Answers (1)

2020-11-07
Given:
\(\displaystyle{9}^{{x}}-{3}^{{{x}+{1}}}+{1}={0}\)
\(\displaystyle{3}^{{{2}{x}}}-{3}^{{{x}}}\times{3}+{1}={0}\)
Let \(\displaystyle{3}^{{x}}\) be u:
\(\displaystyle{u}^{{2}}-{u}\times{3}+{1}={0}\)
\(\displaystyle{u}=\frac{{{3}+\sqrt{{5}}}}{{2}},{u}=\frac{{{3}-\sqrt{{5}}}}{{2}}\)
Substitute back \(\displaystyle{3}^{{x}}\):
\(\displaystyle{3}^{{x}}=\frac{{{3}+\sqrt{{5}}}}{{2}}\)
\(\displaystyle{\ln{{\left({3}^{{x}}\right)}}}={\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}\)
\(\displaystyle{x}{\ln{{\left({3}\right)}}}={\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}\)
\(\displaystyle{x}=\frac{{{\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}}}{{{\ln{{\left({3}\right)}}}}}\)
Solve \(\displaystyle{3}^{{x}}=\frac{{{3}-\sqrt{{5}}}}{{2}}\) similarly:
\(\displaystyle{x}=\frac{{{\ln{{\left(\frac{{{3}-\sqrt{{5}}}}{{2}}\right)}}}}}{{{\ln{{\left({3}\right)}}}}}\)
0

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