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

If two lines with slope $m_1\&m_2$ intersect such that $\theta$ is the angle between them, then

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|⟶\left(1\right)$

The two curves,

$y=a^x$ and $y=b^x$ (where $a\ne b$)

Intersect for $x=0$ at $(0,1)$

Now,

Slope of the tangent at $(0,1)$ to the curve $y=a^x$ is $m_1=\frac{d}{dx}{a}^x{]}_{(0,1)}={loga}^x{]}_{(0,1)}=loga$

Slope of the tangent at $(0,1)$ to the curve $y=b^x$ is $m_2=\frac{d}{dx}{b}^x{]}_{(0,1)}=logb{b^x]}_{0,1}=logb$

$\therefore$ the angle between them is obtained using $\left(1\right)$

$tan\alpha =\left|\frac{loga-logb}{1+log.logb}\right|$

Hence, this is the answer.