Solve the equation.

x ² 3 ^x −  4( 3 ^x) = 0    ?

$$\\x^2 3^x-4(3^x)=0 \\ 3^x ( x^2-4 ) =0 \\ 3^x ( x-2 ) ( x+2 ) =0\\\\ \underbrace{3^x}_{=0} \times( \underbrace{x-2}_{=0} ) \times( \underbrace{x+2}_{=0} ) =0 \\\\ \boxed{3^x=0} \quad | \quad \ln{} \\\\ \ln{(3^x)} = \ln{(0)} \\ x\ln{(3)} = \ln{(0)} \\ x= { \ln{(0)} \over \ln{(3)} } } \quad | \quad \ln{(0)} \mbox{ no solution !}\\\\ \boxed{x-2=0} \quad \Rightarrow \quad \boxed{x=x_1=2}\\\\ \boxed{x+2=0} \quad \Rightarrow \quad \boxed{x=x_2=-2}$$

e^2x − 9e^x + 8 = 0  ?

  $$\\e^{2x}=e^xe^x \quad | \quad \mbox{ set }\quad \boxed{z=e^x}\\ z^2 -9z +8=0\\ \underbrace{1}_{a=1}z^2 \underbrace{-9}_{b=-9}z \underbrace{+8}_{c=8} =0\\ \boxed{z_{1,2}= { -b\pm\sqrt{b^2-4ac} \over 2a } }\\\\ z_{1,2}= { 9\pm\sqrt{81-4*1*8} \over 2*1 } \\\\ z_{1,2}= { 9\pm\sqrt{81-32} \over 2} \\\\ z_{1,2}= { 9\pm7\over 2} \\\\ \boxed{z_1= 8 \quad z_2 = 1} \\\\ e^x=z \quad | \quad \ln\\\\ \ln{(e^x)}=\ln{(z)}\\\\ x\ln{(e)}=\ln{(z)} \quad | \quad \ln{(e)}= 1 \quad !\\\\ \boxed{x=\ln{(z)}} \\\\$$

$$\\x=x_1=\ln{(8)}=2.07944154168\\ x=x_2=\ln{(1)}=0\\ \boxed{x_1=2.07944154168 \qquad x_2=0}$$

e^6x + 5e^3x − 14 = 0 ?

$$\\e^{6x}=e^{3x}e^{3x} \quad | \quad \mbox{ set }\quad \boxed{z=e^{3x}}\\ z^2 +5z -14=0\\ \underbrace{1}_{a=1}z^2 \underbrace{+5}_{b=5}z \underbrace{-14}_{c=-14} =0\\ \boxed{z_{1,2}= { -b\pm\sqrt{b^2-4ac} \over 2a } }\\\\ z_{1,2}= { -5\pm\sqrt{25-4*1*(-14)} \over 2*1 } \\\\ z_{1,2}= { -5\pm\sqrt{25+56} \over 2} \\\\ z_{1,2}= { -5\pm9\over 2} \\\\ \boxed{z_1= 2 \quad z_2 = -7} \\\\ e^{3x}=z \quad | \quad \ln\\\\ \ln{(e^{3x})}=\ln{(z)}\\\\ 3x\ln{(e)}=\ln{(z)} \quad | \quad \ln{(e)}= 1 \quad !\\\\ \boxed{x={1\over 3}\ln{(z)}} \\\\$$

$$\\x=x_1={ \ln{(2)}\over 3} } =0.23104906019\\ x=x_2={ \ln{(-7)}\over 3} } = \mbox{no solution !}\\\\ \boxed{x=0.23104906019}$$

Really nice work Heureka!

I'd give you more points if I could.