3^(2x)+9=10*3^x
\(\begin{array}{rcll} 3^{2x}+9 &=& 10\cdot 3^x \qquad & | \qquad - 10\cdot 3^x \\ 3^{2x}- 10\cdot 3^x + 9 &=& 0 \\ 3^{x+x}- 10\cdot 3^x + 9 &=& 0 \\ 3^x \cdot 3^x - 10\cdot 3^x + 9 &=& 0 \qquad & | \qquad z= 3^x\\ z\cdot z - 10\cdot z + 9 &=& 0 \\ z^2 - 10\cdot z + 9 &=& 0 \\ \end{array}\)
\( \boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~} \)
\(\begin{array}{rcll} z^2 - 10\cdot z + 9 &=& 0 \quad | \quad a=1 \quad b = -10 \quad c = 9 \\ z &=& \dfrac{-(-10) \pm \sqrt{(-10)^2-4\cdot 1 \cdot 9} }{2\cdot 1} \\ z &=& \dfrac{ 10 \pm \sqrt{100-36} }{2} \\ z &=& \dfrac{ 10 \pm \sqrt{64} }{2} \\ z &=& \dfrac{ 10 \pm 8 }{2} \\ z &=& 5 \pm 4 \\\\ z_1 &=& 5+4 \\ \mathbf{z_1} &\mathbf{=}& \mathbf{9} \\\\ z_2 &=& 5-4 \\ \mathbf{z_2} &\mathbf{=}& \mathbf{1} \end{array}\)
\(\begin{array}{rcll} z &=& 3^x \qquad & | \qquad \ln{()} \\ \ln{(z)} &=& \ln{(3^x)} \\ \ln{(z)} &=& x \cdot \ln{(3)} \\ x &=& \frac{ \ln{(z)} }{ \ln{(3)} } \\\\ x_1 &=& \frac{ \ln{(z_1)} }{ \ln{(3)} } \\ x_1 &=& \frac{ \ln{(9)} }{ \ln{(3)} } \\ x_1 &=& \frac{ \ln{(3^2)} }{ \ln{(3)} } \\ x_1 &=& \frac{ 2 \ln{(3)} }{ \ln{(3)} } \\ \mathbf{x_1} &\mathbf{=}& \mathbf{2} \\\\ x_2 &=& \frac{ \ln{(z_2)} }{ \ln{(3)} } \\ x_2 &=& \frac{ \ln{(1)} }{ \ln{(3)} } \qquad & | \qquad \ln{(1)} = 0\\ x_2 &=& \frac{ 0 }{ \ln{(3)} } \\ \mathbf{x_2} &\mathbf{=}& \mathbf{0} \\\\ \end{array}\)
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