Log attack

Case (i)

If $a = b$ the equation $a^x + b^x = 1$ where $0< a, b < 1$ and $a + b < 1$ becomes $2a^x =1 $. Then $\log 2 + x \log a = 0$ so the solution can be given in three equivalent forms: $$x=\frac{-\log 2}{\log a}=\frac{\log 0.5}{\log a}=\frac{\log 2}{\log 1/a}.$$ 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: $$4^x/2^x+1=4^x$$ So $$2^x+1=(2^x{)}^2.$$ and the equation becomes: $y + 1= y^2$ or $y^2- y- 1 = 0$. Using the quadratic formula, the negative solution to the quadratic can be ignored as $2^x$ is never negative, so the solution is: $$y=\frac{1+\sqrt{5}}{2}$$ (note this equals the golden ratio, $\phi $). Therefore $2^x = \phi$ and so $x\ln 2 = \ln \phi$ and the solution is: $$x=\frac{\ln \varphi }{\ln 2}=\frac{\ln (1+\sqrt{5})/2}{\ln 2}$$ giving $x = 0.69424$ to 5 significant figures.