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The two curves $y=3^{x}$ and $y=5^{x}$ intersect at an angle
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(\frac{\log 3-\log 5}{1+\log 3 \log 5}\right)$
Given curves are $y=3^{x}$ ...(i)
and $y=5^{x}$ ...(ii)
intersect at the point $(0,1)$
Now, differentiating eqs. (i) and (ii) w.r.t. $x$, we get
$$
\begin{array}{l}
\frac{d y}{d x}=3^{x} \log 3 \text { and } \frac{d y}{d x}=5^{x} \log 5 \\
\Rightarrow\left(\frac{d y}{d x}\right)_{(0,1)}=\log 3 \text { and }\left(\frac{d y}{d x}\right)_{(0,1)}=\log 5 \\
\Rightarrow m_{1}=\log 3 \text { and } m_{2}=\log 5
\end{array}
$$
Angle between these curves is given by
$$
\begin{array}{l}
\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}} \\
\Rightarrow \tan \theta=\frac{\log 3-\log 5}{1+\log 3 \cdot \log 5} \\
\Rightarrow \theta=\tan ^{-1}\left(\frac{\log 3-\log 5}{1+\log 3 \log 5}\right)
\end{array}
$$
and $y=5^{x}$ ...(ii)
intersect at the point $(0,1)$
Now, differentiating eqs. (i) and (ii) w.r.t. $x$, we get
$$
\begin{array}{l}
\frac{d y}{d x}=3^{x} \log 3 \text { and } \frac{d y}{d x}=5^{x} \log 5 \\
\Rightarrow\left(\frac{d y}{d x}\right)_{(0,1)}=\log 3 \text { and }\left(\frac{d y}{d x}\right)_{(0,1)}=\log 5 \\
\Rightarrow m_{1}=\log 3 \text { and } m_{2}=\log 5
\end{array}
$$
Angle between these curves is given by
$$
\begin{array}{l}
\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}} \\
\Rightarrow \tan \theta=\frac{\log 3-\log 5}{1+\log 3 \cdot \log 5} \\
\Rightarrow \theta=\tan ^{-1}\left(\frac{\log 3-\log 5}{1+\log 3 \log 5}\right)
\end{array}
$$
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