Question

If the angle between the curves $y=2^x$ and $y=3^x$ is $\alpha ,$ then the value of $\tan \alpha$ is equal to :

• A. $\frac{\log \left(\frac{3}{2}\right)}{1+\left(\log 2\right)\left(\log 3\right)}$
• B. $\frac{6}{7}$
• C. $\frac{1}{7}$
• D. $\frac{\log \left(6\right)}{1+\left(\log 2\right)\left(\log 3\right)}$
• E. $0^o$

Answer 1

The correct option is A $\frac{\log \left(\frac{3}{2}\right)}{1+\left(\log 2\right)\left(\log 3\right)}$

Given curves are $y=2^x$ and $y=3^x$
The point of intersection is $3^x=2^x\Rightarrow x=0$
On differentiating w.r.t. $x,$ we get
$\frac{dy}{dx}=2^x\log 2=m_1$
and $\frac{dy}{dx}=3^x\log 3=m_2$
Therefore, $\tan \alpha =\frac{m_2-m_1}{1+m_1{m}_2}$
$=\frac{3^x\log 3-2^x\log 2}{1+3^x\times 2^x\log 3\times \log 2}$
At $x=0$,

$\tan \alpha =\frac{3^0\log 3-2^0\log 2}{1+3^0\times 2^0\log 2\log 3}$
$=\frac{\log \frac{3}{2}}{1+\log 2\log 3}$