Question

If the curves $y=a^x$ and $y=b^x$ intersects at angle $\alpha$, then it is given by

• A. $\tan \alpha =\left|\frac{\log \left(ab\right)}{1-\log \left(ab\right)}\right|$
• B. $\tan \alpha =\left|\frac{\log \left(a/b\right)}{1+\log \left(a\right)\log \left(b\right)}\right|$
• C. $\tan \alpha =\left|\frac{\log \left(ab\right)}{1+\log \left(ab\right)}\right|$
• D. None

Answer 1

The correct option is B $\tan \alpha =\left|\frac{\log \left(a/b\right)}{1+\log \left(a\right)\log \left(b\right)}\right|$

$po\mathrm{int}\text{of intersect of curves}$
${\text{a}}^x=b^x$
$\Rightarrow {\left(\frac{a}{b}\right)}^x=1$
$x=0$
$Hencepo\mathrm{int}of\mathrm{int}er\sec tion:(0,1)$

$y=a^x$
$\frac{dy}{dx}=a^x\ln a$
${\frac{dy}{dx}}_{(0,1)}=m_1=\ln a$

$y=b^x$
$\frac{dy}{dx}=b^x\ln b$
${\frac{dy}{dx}}_{(0,1)}=m_2=\ln b$

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$plugginginthevalues$
$\tan \theta =\left|\frac{\log a-\ln b}{1+\left(\log a\right)\left(\log b\right)}\right|$
$=\left|\frac{\log \left(\frac{a}{b}\right)}{1+\left(\log a\right)\left(\log b\right)}\right|$