Question

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

• A. $\tan \theta = \frac{\log(3/7)}{1 + (\log 3)(\log 7)}$
• B. $\tan \theta = \frac{\log(7/3)}{1 + (\log 3)(\log 7)}$
• C. $\tan \theta = \frac{\log(3/7)}{1 - (\log 3)(\log 7)}$
• D. $\tan \theta = \frac{\log(7/3)}{1 - (\log 3)(\log 7)}$

Hint:

The derivative of $a^x$ is $a^x \log_e a$. Do not confuse this with the power rule! A common mistake is writing $x \cdot a^{x-1}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The curves $y = 3^x$ and $y = 7^x$ intersect where $3^x = 7^x$, which only occurs at $x = 0$. At $x = 0, y = 1$. The angle between curves is the angle between their tangents at the point of intersection $(0, 1)$.

Step 2: Formula Application:
The slope of the tangent $m = \frac{dy}{dx}$. For $y = a^x$, $\frac{dy}{dx} = a^x \log a$. The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.

Step 3: Explanation:
For $y = 3^x$, $m_1 = 3^0 \log 3 = \log 3$. For $y = 7^x$, $m_2 = 7^0 \log 7 = \log 7$. $\tan \theta = \frac{\log 7 - \log 3}{1 + (\log 7)(\log 3)} = \frac{\log(7/3)}{1 + (\log 3)(\log 7)}$.

Step 4: Final Answer:
The value is $\tan \theta = \frac{\log(7/3)}{1 + (\log 3)(\log 7)}$.