Question

If the two curves $y=a^{x}andy=b^x$ intersect at $\angle \alpha$, then $tan\alpha$ equal

• A. $\frac{loga-logb}{1+logalogb}$
• B. $\frac{loga+logb}{1-logalogb}$
• C. $\frac{loga-logb}{1-logalogb}$
• D. None of these

Answer 1

The correct option is A $\frac{loga-logb}{1+logalogb}$

Given $y=a^x,y=b^x$

$a^x=b^x⟹x=0$

$\therefore$ curves $y=a^x$ and $y=b^x$ intersects at $x=0,y=1$

Let ${\theta }_1$ be the angle made by the tangent of $y=a^x$ at $x=0$ with $x$ axis

$\therefore$ slope $=m_1=\tan {\theta }_1=\frac{dy}{dx}{|}_{x=0}=a^x\log a{|}_{x=0}$

$=\log a$

Let ${\theta }_2$ be the angle made by the tangent of $y=b^x$ at $x=0$ with $x$ axis

$\therefore$ slope $=m_2=\tan {\theta }_2=\frac{dy}{dx}{|}_{x=0}=b^x\log b{|}_{x=0}$

$=\log b$

Given angle between curves $2\alpha$

${\theta }_1-{\theta }_2=\alpha$

$\therefore \tan \alpha =\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}$

$=\frac{m_1–m_2}{1+m_1{m}_2}$

$\therefore \tan \alpha =\frac{\log a-\log b}{1+\log a\log b}$