If the angle between the curves y = a^x and y | vedclass

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(D) To find the intersection point,set $y = a^x$ and $y = b^x$ equal to each other:$a^x = b^x \implies (a/b)^x = 1 \implies x = 0$.For $x = 0$,$y = a^

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

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(D) To find the intersection point,set $y = a^x$ and $y = b^x$ equal to each other:$a^x = b^x \implies (a/b)^x = 1 \implies x = 0$.For $x = 0$,$y = a^0 = 1$. Thus,the intersection point is $(0, 1)$.Now,find the slopes of the tangents at $(0, 1)$:For $y = a^x$,$\frac{dy}{dx} = a^x \ln a$. At $x = 0$,$m_1 = \ln a$.For $y = b^x$,$\frac{dy}{dx} = b^x \ln b$. At $x = 0$,$m_2 = \ln b$.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 of $m_1$ and $m_2$:$	an \alpha = \left| \frac{\ln a - \ln b}{1 + \ln a \ln b} ight|$.

If the angle between the curves $y = a^x$ and $y = b^x$ is $\alpha$,find the modulus of $\tan \alpha$.

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