if we can match teh bases we can solve
because if x=x and xᵃ=xᵇ, we can conclude that a=b
16=2⁴
32=2⁵
rememeber that \((x^m)^n=x^{mn}\)
\(16^{3x+2}=32^{-2x-7}\)
\((2^4)^{3x+2}=(2^5)^{-2x-7}\)
\(2^{4(3x+2)}=2^{5(-2x-7)}\)
2=2 so we conclude that 4(3x+2)=5(-2x-7)
4(3x+2)=5(-2x-7)
expand/distribute
12x+8=-10x-35
add 10x both sides
22x+8=-35
minus 8 both sides
22x=-43
divide both sides by 22
x=-43/22
