help......4..........

6(4^x+9^x)=13*6^x

$\begin{array}{rcll} 6\cdot (4^x+9^x) &=& 13\cdot 6^x \qquad & | \qquad :6\\ 4^x+9^x &=& \frac{13}{6}\cdot 6^x \qquad & | \qquad :6^x\\ \frac{ 4^x } {6^x} + \frac{ 9^x } {6^x} &=& \frac{13}{6}\\ \left(\frac46 \right)^x + \left(\frac96 \right)^x &=& \frac{13}{6}\\ \left(\frac23 \right)^x + \left(\frac32 \right)^x &=& \frac{13}{6}\\ \left( \frac{1}{\frac32} \right)^x + \left(\frac32 \right)^x &=& \frac{13}{6}\\ \frac{1}{ \left( \frac32\right)^x } + \left(\frac32 \right)^x &=& \frac{13}{6} \qquad & | \qquad z= \left(\frac32 \right)^x\\ \frac{1}{ z } + z &=& \frac{13}{6} \qquad & | \qquad \cdot z\\ \frac{z}{ z } + z^2 &=& \frac{13}{6} \cdot z\\ 1 + z^2 &=& \frac{13}{6} \cdot z\\ z^2 - \frac{13}{6} \cdot z + 1 &=& 0 \\ \end{array}$

$\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}$

$\begin{array}{rcll} z^2 - \frac{13}{6} \cdot z + 1 &=& 0 \quad | \quad a=1 \quad b =-\frac{13}{6} \quad c = 1 \\ z &=& \dfrac{-(-\frac{13}{6}) \pm \sqrt{(-\frac{13}{6})^2-4\cdot 1 \cdot 1} }{2\cdot 1} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{13^2}{6^2}-4 } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{13^2-4\cdot 6^2}{6^2} } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{169-144}{6^2} } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{25}{6^2} } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \sqrt{ \frac{5^2}{6^2} } }{2} \\ z &=& \dfrac{ \frac{13}{6} \pm \frac{5}{6} }{2} \\ z &=& \frac{13}{12} \pm \frac{5}{12} \\ z_1 &=& \frac{13}{12} + \frac{5}{12} \\ z_1 &=& \frac{18}{12} \\ \mathbf{z_1} &\mathbf{=}& \mathbf{\frac{3}{2}} \\\\ z_2 &=& \frac{13}{12} - \frac{5}{12} \\ z_2 &=& \frac{8}{12} \\ \mathbf{z_2} &\mathbf{=}& \mathbf{\frac{2}{3}} \end{array}$

$\begin{array}{rcll} z&=& \left(\frac32 \right)^x \qquad & | \qquad \ln{()} \\ \ln{(z)} &=& x \cdot \ln{ \left(\frac32 \right) } \\ x &=& \frac{ \ln{(z)} }{ \ln{ \left(\frac32 \right) } } \\\\ x_1 &=& \frac{ \ln{(z_1)} }{ \ln{ \left(\frac32 \right) } } \\ x_1 &=& \frac{ \ln{\left(\frac32 \right)} }{ \ln{ \left(\frac32 \right) } } \\ \mathbf{x_1} &\mathbf{=}& \mathbf{1} \\\\ x_2 &=& \frac{ \ln{(z_2)} }{\ln{ \left(\frac32 \right) } } \\ x_2 &=& \frac{ \ln{ \left(\frac23 \right) } }{ \ln{ \left(\frac32 \right) } } \\ x_2 &=& \frac{ \ln{ \left( \frac{1}{ \frac32} \right) } }{ \ln{ \left(\frac32 \right) } } \\ x_2 &=& \frac{ \ln{(1)} - \ln{ \left( \frac32 \right) } }{ \ln{ \left(\frac32 \right) } } \qquad & | \qquad \ln{(1)} = 0\\ x_2 &=& \frac{ 0 - \ln{ \left( \frac32 \right) } }{ \ln{ \left(\frac32 \right) } } \\ x_2 &=& \frac{ -\ln{ \left( \frac32 \right) } }{ \ln{ \left(\frac32 \right) } } \\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1} \\\\ \end{array}$