answer please....

6^(2x+4)=3^(3x)*2^(x+8)

$\begin{array}{rcll} 6^{2x+4} &=& 3^{3x}\cdot 2^{x+8} \\ 6^{2x}\cdot 6^4 &=& 3^{3x}\cdot 2^x \cdot 2^8 \\ 6^{2x}\cdot (2\cdot 3)^4 &=& 3^{3x}\cdot 2^x \cdot 2^8 \\ 6^{2x}\cdot 2^4 \cdot 3^4 &=& 3^{3x}\cdot 2^x \cdot 2^8 \qquad & | \qquad : 2^4\\ 6^{2x}\cdot 3^4 &=& 3^{3x}\cdot 2^x \cdot 2^4 \\ (2\cdot 3)^{2x}\cdot 3^4 &=& 3^{3x}\cdot 2^x \cdot 2^4 \\ 2^{2x}\cdot 3^{2x}\cdot 3^4 &=& 3^{3x}\cdot 2^x \cdot 2^4 \qquad & | \qquad : 2^x\\ 2^{2x-x}\cdot 3^{2x}\cdot 3^4 &=& 3^{3x} \cdot 2^4 \\ 2^x\cdot 3^{2x}\cdot 3^4 &=& 3^{3x} \cdot 2^4 \qquad & | \qquad : 3^{2x}\\ 2^x \cdot 3^4 &=& 3^{3x-2x} \cdot 2^4 \\ 2^x \cdot 3^4 &=& 3^x \cdot 2^4 \\ \frac{2^x} {3^x} &=& \frac{ 2^4 } {3^4} \\ \left(\frac23 \right)^x &=& \left(\frac23 \right)^4 \\ \mathbf{ x } & \mathbf{=} & \mathbf{4} \\ \end{array}$