# At what angle $\theta$ do the curves $y=a^{x}$ and $y=b^{x}$ intersect $(a{\neq}b)$?

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) • If the curves y=f(x),y=g(x) intersect at (x_1,y_1) then the angle of intersection of the curves is the angle between the tangents at the point (x_1,y_1) • If m_1=t'(x_1) and m_2=g'(x_1) then the angle is \tan ^{-1} \large\frac{m_1-m_2}{1+m_1m_2} when the curves are orthogonal , the tangents are perpendicular to each other and m_1m_2=-1 Step 1: Let y=a^x----(i) and y=b^n ----(ii) intersect at (x_1,y_1) Step 2: \large\frac{dy}{dx}$$=a^x \log a$ and $\large\frac{dy}{dx}$$=b ^x \log b$ on differentiating (i) and (ii)
Step 3:
The slopes of the tangents to the curves at $(x_1,y_1)$ are $m_1=a^{x_1} \log b$ and $m_2=b^{x_1} \log b$
The angle between the curves is
$\tan ^{-1} \large\frac{m_1-m_2}{1+m_1m_2}=\frac{a^{x_1}\log a -b^{x_1}\log b}{1+ a ^{x_1} \log a b^{x_1}\log b}$
Step 4:
Now $y_1=a^{x_1}$ and $y_1=b^{x_1}$
Therefore the angle is $\tan ^{-1} \large\frac{y_1 \log a-y_1 \log b}{1+{y_1}^2 \log a \log b}$
$a^{x_1}=b^{x_1}$ but $a \neq b=>x_1=0$
Therefore $y_1=1$
The angle is $\tan ^{-1} \large\frac{\log a-\log b}{1+\log a \log b}$