Question

The angle between the curves $y=a^x$ of and $y=b^x$ is equal to

• A. ${\tan }^{-1}\left(\left|\frac{a-b}{1+ab}\right|\right)$
• B. ${\tan }^{-1}\left(\left|\frac{a+b}{1-ab}\right|\right)$
• C. ${\tan }^{-1}\left(\left|\frac{\log b+\log a}{1+\log a\log b}\right|\right)$
• D. ${\tan }^{-1}\left(\left|\frac{\log a+\log b}{1-\log a\log b}\right|\right)$
• E. ${\tan }^{-1}\left(\left|\frac{\log a-\log b}{1+\log a\log b}\right|\right)$

Answer 1

Answer: (E)

The point of intersection of given curves is (0, 1).
On differentiating given curves, we get
$\frac{dy}{dx}=a^x\log a,\frac{dy}{dx}=b^x\log b$
$\Rightarrow$ $m_1=a^x\log a,m_2=b^x\log b$
At (0, 1), $m_1=loga,m_2=logb$
$\therefore$ $\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$
$\Rightarrow$ $\theta ={\tan }^{-1}\left(\frac{\log a-\log b}{1+\log a\log b}\right)$