If a = b the equation a^x + b^x = 1 where 0 < a, b < 1 and a + b < 1 becomes 2a^x = 1 . Then log2 + x loga = 0 so the solution can be given in three equivalent forms:

x = -log2 loga = log0.5 loga = log2 log1/a .

Note we can use natural logarithms here or logarithms to any base.

Case (ii)

Solve a^x + b^x = 1, where a=1/2 and b=1/4. The equation is:

(1/2)x + (1/4)x = 1,

Here you could substitute y = (1/2)^x but Trevor multiplied by 4^x and used the substitution y = 2^x. By Trevor's method:

4x/2x + 1 = 4x

So

2x + 1 = (2x)2.

and the equation becomes: y + 1 = y² or y²- y- 1 = 0. Using the quadratic formula, the negative solution to the quadratic can be ignored as 2^x is never negative, so the solution is:

y= 1+ Ö5 2

(note this equals the golden ratio, f). Therefore 2^x = f and so xln2 = lnf and the solution is:

x = lnf ln2 = ln(1 + Ö5)/2 ln2

giving x = 0.69424 to 5 significant figures.