Explanation

(b)
The equations of given curves are
$y = a^{x}$ …(i)
And $y = b^{x}$ …(ii)
From Eq. (i) $m_{1} = \frac{dy}{dx} = a^{x}\log a$
And from Eq. (ii) $m_{2} = \frac{dy}{dx} = b^{x}\log b$
From Eqs. (i) and (ii), we get
$a^{x} = b^{x} \Rightarrow x = 0$
Let $\alpha$ be the angle at which the two curves intersect
$\therefore\ \tan\alpha = \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}}$
$= \frac{a^{x}\log a - b^{x}\log b}{1 + a^{x}b^{x}\left( \log a \right)\left( \log b \right)}$
$= \frac{\log\frac{a}{b}}{1 + \left( \log a \right)(\log b)}$