Question

If the curve $y=a^x$ and $y=b^x$ intersect at angle $\alpha$ then, $\tan \alpha =$

• A. $\frac{a-b}{1+ab}$
• B. $\frac{\log a-\log b}{1+\log a\log b}$
• C. $\frac{a+b}{1-ab}$
• D. $\frac{\log a+\log b}{1-\log a\log b}$

Answer 1

The correct option is B

$\frac{\log a-\log b}{1+\log a\log b}$

Explanation for the correct option.

Given, curve $y=a^x$ and $y=b^x$ intersect at an angle $\alpha$.

We know that angle between curves is nothing but the angle between their tangents.

If two lines with slopes ${\mathrm{m}}_1\&{\mathrm{m}}_2$ intersect such that $\theta$ is the angle between them, then
$\tan \theta =\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|....\left(1\right)$
The two curves,$y=a^x$ and $y=b^x$ (where $a\equiv b$)
Intersect for $x=0$ at $(0,1)$
Now, the slope of the tangent at $(0,1)$ to the curve $y=a^x$ is:

$\begin{array}{rcl}\therefore m_1& =& \frac{d}{dx}{a}^x{|}_{(0,1)}\\ & =& \log a^x{|}_{(0,1)}\end{array}$

The slope of the tangent at $(0,1)$ to the curve $y=b^x$ is

$\begin{array}{rcl}\therefore m_2& =& \frac{d}{dx}{b}^x{|}_{\left(0,1\right)}\\ & =& \log b^x{|}_{(0,1)}\end{array}$

$=\log b$
$\therefore$ the angle between them is :$\tan \alpha =\left|\frac{\log a-\log b}{1+\log ·\log b}\right|$.

Hence, option B is correct.