If the two curves y=a^x and y=b^x intersect a | vedclass

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(A) Given curves are $y=a^x$ and $y=b^x$. At the point of intersection,$a^x = b^x$,which implies $x=0$. Thus,the curves intersect at the point $(0, 1)

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

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(A) Given curves are $y=a^x$ and $y=b^x$. At the point of intersection,$a^x = b^x$,which implies $x=0$. Thus,the curves intersect at the point $(0, 1)$. Let $m_1$ be the slope of the tangent to $y=a^x$ at $x=0$. $m_1 = \left. \frac{dy}{dx} ight|_{x=0} = \left. a^x \ln a ight|_{x=0} = \ln a$. Let $m_2$ be the slope of the tangent to $y=b^x$ at $x=0$. $m_2 = \left. \frac{dy}{dx} ight|_{x=0} = \left. b^x \ln b ight|_{x=0} = \ln b$. The angle $\alpha$ between the curves is given by $	an \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. Substituting the values,we get $	an \alpha = \frac{\ln a - \ln b}{1 + \ln a \ln b}$.

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

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