Can someone please help me with this??

Answer:
Step-by-step explanation:
[tex]12.\ a^v=1\to a^v=a^0\to v=0\\\\\text{because}\ a^0=1\ \text{for any}\ a\neq0.\\\\13.\ \ b^{-w}=\dfrac{1}{b^9}\to \dfrac{1}{b^w}=\dfrac{1}{b}\to b=9\\\\\text{because}\ a^{-n}=\dfrac{1}{a^n}\ \text{for any}\ a\neq0\\\\14.\ \dfrac{3^3}{3^y}=\dfrac{1}{9}\to\dfrac{3^3}{3^y}=\dfrac{3^0}{3^2}\to 3-y=-2\to y=5\\\\\text{used}\ \dfrac{a^m}{a^n}=a^{n-m}\\\\15.\ 5^8\cdot5^z=1\to5^{8+z}=5^0\to8+z=0\to z=-8\\\\\text{used}\ a^n\cdot a^m=a^{n+m}[/tex]
[tex]16.\ 7^n\cdot 7^m=1\to7^{n+m}=7^0\to n+m=0\to n=-m\\\\17.\ \dfrac{3^x\cdot2^2}{3^4}=\dfrac{4}{3}\to3^{x-4}\cdot2^2=3^{-1}\cdot2^2\to x-4=-1\to x=3[/tex]