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

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(B) The curves are $y = a^x$ and $y = b^x$. Setting $a^x = b^x$,we get $(a/b)^x = 1$,which implies $x = 0$. At $x = 0$,$y = a^0 = 1$. So,the point of

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

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(B) The curves are $y = a^x$ and $y = b^x$. Setting $a^x = b^x$,we get $(a/b)^x = 1$,which implies $x = 0$. At $x = 0$,$y = a^0 = 1$. So,the point of intersection is $(0, 1)$. For the first curve $y = a^x$,the slope of the tangent is $m_1 = \frac{dy}{dx} = a^x \ln a$. At $(0, 1)$,$m_1 = a^0 \ln a = \ln a$. For the second curve $y = b^x$,the slope of the tangent is $m_2 = \frac{dy}{dx} = b^x \ln b$. At $(0, 1)$,$m_2 = b^0 \ln b = \ln b$. The angle $\alpha$ between the curves is given by $	an \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. Substituting the values,$	an \alpha = \frac{\ln a - \ln b}{1 + \ln a \ln b} = \frac{\log a - \log b}{1 + \log a \log b}$.

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

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