Question

The two curves \(y=3^x\) and \(y=5^x\) intersect at an angle

• A. \(\tan^{-1}\left(\frac{\log 3-\log 5}{1+\log 3\log 5}\right)\)
• B. \(\tan^{-1}\left(\frac{\log 3+\log 5}{1-\log 3\log 5}\right)\)
• C. \(\tan^{-1}\left(\frac{\log 3+\log 5}{1+\log 3\log 5}\right)\)
• D. \(\tan^{-1}\left(\frac{\log 3-\log 5}{1-\log 3\log 5}\right)\)

Hint:

Angle between curves is angle between tangents: \(\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|\). Compute slopes at intersection point.

Answer 1

The Correct Option is A

Step 1: Find point of intersection.
\[ 3^x=5^x \Rightarrow \left(\frac{3}{5}\right)^x=1 \Rightarrow x=0 \] Then:
\[ y=3^0=1 \] Intersection point is \((0,1)\).
Step 2: Find slopes of tangents at intersection.
For \(y=3^x\):
\[ \frac{dy}{dx}=3^x\ln 3 \Rightarrow m_1=\ln 3 \ \text{at}\ x=0 \] For \(y=5^x\):
\[ \frac{dy}{dx}=5^x\ln 5 \Rightarrow m_2=\ln 5 \ \text{at}\ x=0 \] Step 3: Angle between curves formula.
\[ \tan\theta = \left|\frac{m_2-m_1}{1+m_1m_2}\right| \] Substitute:
\[ \tan\theta=\left|\frac{\ln 5-\ln 3}{1+\ln 3\ln 5}\right| \] Using base-10 logs (\(\ln a = 2.303\log a\)), constant cancels, so:
\[ \theta=\tan^{-1}\left(\frac{\log 3-\log 5}{1+\log 3\log 5}\right) \] Final Answer: \[ \boxed{\tan^{-1}\left(\frac{\log 3-\log 5}{1+\log 3\log 5}\right)} \]