Question

If the two curves $y=a^x$ and $y=b^x$ intersect at an angle $\alpha$, then tan $\alpha$ equals.

• A. $\frac{\log a-\log b}{1+\log a\log b}$
• B. $\frac{\log a+\log b}{1-\log a\log b}$
• C. $\frac{\log a-\log b}{1-\log a\log b}$
• D. none of these

Answer 1

Answer: (A)

For the curve $y=a^x,\frac{dy}{dx}=a^{x}loga$
For the curve $y=b^x,\frac{dy}{dx}=b^{x}logb$
The given curves meet at $\left(1,1\right)$
$\left[\because a^x=b^x\Rightarrow x=0\right]$
At $\left(0,1\right)$
$m_1=\frac{dy}{dx}$ for first curve $=loga$
$m_2=\frac{dy}{dx}$ for first curve $=logb$
$\therefore tan\alpha =\frac{m_1-m_2}{1+m_1{m}_2}$$=\frac{loga-logb}{1+logalogb}$