(1) Can also be written ex=e-2x+1, e3x=1+e2x.
Let y=ex, then y3=1+y2, y3-y2-1=0.
Let f(y)=y3-y2-1, then f'(y)=3y2-2y. Newton's iterative Method:
yn+1=yn-f(yn)/f'(yn), where y0=1 (arbitrarily).
y1=1-(-1)/(1)=2; y2=2-3/8=13/8; y3=1.485...; y4=1.465...; etc.
y stabilises at y=1.46557123188 approx.
ex=1.46557123188, x=ln(1.46557123188)=0.382245 approx.
(2) Can also be interpreted ex=e-2(x+1), which is much easier to solve, because the exponents are equal:
x=-2x-2, 3x=-2, x=-⅔.