(3ˣ)ˣ×2x=(3^x²)×2x.
x must be negative to yield a negative product and x must be fractional since -1/15 is a fraction. x² is small and positive, close to zero, so 3^x² is nearly 3⁰=1.
Therefore 2x=-1/15, x=-1/30 is an approximate solution.
We can use Newton’s Method to get a more accurate result.
Let f(x)=(3^x²)×2x+1/15. 3^x²=e^(x²ln(3)).
f'(x)=2(3^x²)+4x²ln(3)(3^x²).
Let x₀=-1/30, then we can find successive iterations of x:
x₁=x₀-f(x₀)/f'(x₀)=-0.03329276785,
x₂=-0.03329276767,
x₃=-0.0332927676681.
To 6 decimal places x=-0.033293.