To solve for z, we proceed as follows:
[tex](\frac{1}{4})^{3z-1}=16^{z+2}\times64^{z-2}[/tex]Now, we simplify the expression on the right-hand side of the equation as follows:
[tex](\frac{1}{4})^{3z-1}=(4^2)^{z+2}\times(4^3)^{z-2}[/tex][tex](\frac{1}{4})^{3z-1}=4^2^{(z+2)}\times4^3^{(z-2)}[/tex][tex](\frac{1}{4})^{3z-1}=4^{2z+4}\times4^{3z-6}[/tex][tex](\frac{1}{4})^{3z-1}=4^{2z+4+3z-6}[/tex][tex](\frac{1}{4})^{3z-1}=4^{5z-2}[/tex]Now, we simplify the expression on the left-hand side of the equation as follows:
[tex](4^{-1})^{3z-1}=4^{5z-2}[/tex][tex]4^{-1(3z-1)}=4^{5z-2}[/tex][tex]4^{-3z+1}=4^{5z-2}[/tex]Now, since we have both expressions on the left and right hand sides to have a base of 4, we can simply equate their indices, as follow:
[tex]\begin{gathered} 4^{-3z+1}=4^{5z-2} \\ \Rightarrow-3z+1=5z-2 \\ \end{gathered}[/tex]Now, we collect like terms:
[tex]-3z-5z=-2-1[/tex][tex]\begin{gathered} -8z=-3 \\ \Rightarrow z=\frac{-3}{-8} \\ \Rightarrow z=\frac{3}{8} \end{gathered}[/tex]