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:

${\displaystyle x=\frac{-\log 2}{\log a}=\frac{\log 0.5}{\log a}=\frac{\log 2}{\log 1/a}.}$

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:

${\displaystyle (1/2{)}^x+(1/4{)}^x=1,}$

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:

${\displaystyle 4^x/2^x+1=4^x}$

So

${\displaystyle 2^x+1=(2^x{)}^2.}$

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:

${\displaystyle y=\frac{1+\sqrt{5}}{2}}$

(note this equals the golden ratio, $\varphi$). Therefore $2^x=\varphi$ and so $x\ln 2=\ln \varphi$ and the solution is:

${\displaystyle x=\frac{\ln \varphi }{\ln 2}=\frac{\ln (1+\sqrt{5})/2}{\ln 2}}$

giving $x=0.69424$ to 5 significant figures.