Answer:
Step-by-step explanation:
[tex]\left(\frac{2}{9}\right)^3\cdot\left(\frac{2}{9}\right)^8=\left(\frac{2}{9}\right)^{2x-1}\\\left(\frac{2}{9}\right)^{3+8}=\left(\frac{2}{9}\right)^{2x-1}\\\left(\frac{2}{9}\right)^{11}=\left(\frac{2}{9}\right)^{2x-1}\\11=2x-1,\quad x=6[/tex]
