The angle theta,at which the curves y=3^x and | vedclass

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(D) The curves are $y_1 = 3^x$ and $y_2 = 7^x$. They intersect where $3^x = 7^x$,which implies $x=0$. At $x=0$,$y=3^0=1$. So the point of intersection

Vedclass · Accepted Answer

$	an 	heta=\frac{\log \left(\frac{7}{3}ight)}{1+(\log 3)(\log 7)}$

Vedclass · Answer

(D) The curves are $y_1 = 3^x$ and $y_2 = 7^x$. They intersect where $3^x = 7^x$,which implies $x=0$. At $x=0$,$y=3^0=1$. So the point of intersection is $(0, 1)$. The slopes of the tangents are $m_1 = \frac{dy_1}{dx} = 3^x \ln 3$ and $m_2 = \frac{dy_2}{dx} = 7^x \ln 7$. At $(0, 1)$,$m_1 = \ln 3$ and $m_2 = \ln 7$. The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$. $	an 	heta = |\frac{\ln 7 - \ln 3}{1 + (\ln 3)(\ln 7)}| = \frac{\ln(7/3)}{1 + (\ln 3)(\ln 7)}$. Since $\log x$ usually denotes $\log_{10} x$ and $\ln x$ denotes $\log_e x$,the expression is equivalent to $\frac{\log(7/3)}{1 + (\log 3)(\log 7)}$ if the base is consistent.

The angle $\theta$,at which the curves $y=3^x$ and $y=7^x$ intersect,is given by

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