(i) If

2 a^{x} = 1

then

\log 2 + x \log a = 0

x = \frac{\log 2}{\log 1 / a} .

(ii) To solve

{\frac{1}{2}}^{x} + {\frac{1}{4}}^{x} = 1

write

y = (\frac{1}{2})^{x}

then

y^{2} + y = 1

y = \frac{- 1 \pm \sqrt{5}}{2}

. It follows that

x \log 0.5 = \log \frac{\sqrt{5} - 1}{2}

giving

x = 0.69424

to 5 significant figures.
Note that this method introduces an extraneous root of the quadratic equation because

y = (\frac{1}{2})^{x}

is always positive.

(iii) Solving ${\frac{1}{2}}^{x} + {\frac{1}{3}}^{x} = 1$ , numerical approximation methods give $x = 0.7878849$ to seven sig. figs.

Using the Newton Raphson method: $f (x) = a^{x} + b^{x} - 1$ , $f' (x) = a^{x} \ln a + b^{x} \ln b$ and

$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f' (x_{n})} .$