bantuin ya........................

[tex]( \frac{1}{16})^{(3n+1)}=8\times4^n \\ \\ ( \frac{1}{2^4})^{(3n+1)}=2^3\times(2^2)^n \\ \\ 2^{((-4)(3n+1))}=2^{(3+2n)}.............a^{f(n)}=a^{g(n)} \to f(n)=g(n) \\ \\ (-4)(3n+1)=3+2n \\ -12n-4=3+2n \\ -12n-2n=3+4 \\ -14n=7 \\ \\ n= -\frac{7}{14} \\ \\ n=- \frac{1}{2}[/tex]

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