Given:
[tex]4\log _2y+\frac{1}{3}\log _2x^3[/tex][tex]\log a^b=b\log a[/tex]
Then:
[tex]\begin{gathered} 4\log _2y+\frac{1}{3}\log _2x^3 \\ =\log _2y^4+\log _2(x^3)^{\frac{1}{3}} \\ =\log _2y^4+\log _2x \end{gathered}[/tex][tex]\log ab=\log a+\log b[/tex][tex]\begin{gathered} =\log _2y^4+\log _2x \\ =\log _2(y^4x) \end{gathered}[/tex]
