Answer:
[tex]\large\boxed{x^\frac{4}{5}}[/tex]
Step-by-step explanation:
[tex]x^\frac{2}{3}\cdot x^\frac{2}{15}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=x^{\frac{2}{3}+\frac{2}{15}}\qquad\text{LDC = 15}\\\\=x^{\frac{2\cdot5}{3\cdot5}+\frac{2}{15}}=x^{\frac{10}{15}+\frac{2}{15}}=x^{\frac{12}{15}}=x^{\frac{12:3}{15:3}}=x^\frac{4}{5}[/tex]
