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 Question:
We must find the angle of intersection between two exponential curves. The angle of intersection between two curves is defined as the angle between their tangent lines exactly at their point of intersection.

Step 2: Key Formula or Approach:
1. Find the point of intersection by setting the equations equal to each other.
2. Find the slopes ($m_1$ and $m_2$) of the tangent lines by evaluating the derivatives at the intersection point.
3. Use the angle formula: $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.

Step 3: Detailed Explanation:
First, find the point of intersection:
$$3^x = 7^x \implies \left(\frac{3}{7}\right)^x = 1 \implies x = 0$$
When $x = 0$, $y = 3^0 = 1$. The intersection point is $(0, 1)$.
Next, find the derivatives:
For $y = 3^x$, $y' = 3^x \log_e 3$.
At $x = 0$, $m_1 = 3^0 \log 3 = \log 3$.
For $y = 7^x$, $y' = 7^x \log_e 7$.
At $x = 0$, $m_2 = 7^0 \log 7 = \log 7$.
Now, apply the angle formula:
$$\tan \theta = \left| \frac{\log 7 - \log 3}{1 + (\log 3)(\log 7)} \right|$$
Since $7 > 3$, $\log 7 > \log 3$, making the numerator positive. We can drop the absolute value signs and apply the logarithm quotient rule:
$$\tan \theta = \frac{\log(7/3)}{1 + (\log 3)(\log 7)}$$

Step 4: Final Answer:
The angle is given by option (b).