( 16 a _ { n } = 3 ^ { n } - 2
Solve for a_n
a_{n}=\frac{3^{n}-2}{16}
Solve for n
n=\log_{3}\left(16a_{n}+2\right)
a_{n}>-\frac{1}{8}
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16a_{n}=3^{n}-2
The equation is in standard form.
\frac{16a_{n}}{16}=\frac{3^{n}-2}{16}
Divide both sides by 16.
a_{n}=\frac{3^{n}-2}{16}
Dividing by 16 undoes the multiplication by 16.
a_{n}=\frac{3^{n}}{16}-\frac{1}{8}
Divide 3^{n}-2 by 16.
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