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

Complex numbers
Solve the equation:
$$\displaystyle{9}^{{x}}-{3}^{{{x}+{1}}}+{1}={0}$$

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:
$$\displaystyle{x}=\frac{{{\ln{{\left(\frac{{{3}-\sqrt{{5}}}}{{2}}\right)}}}}}{{{\ln{{\left({3}\right)}}}}}$$