Question

The angle between the curves $y=a^x$ and $y=b^x$ is equal to

• A. ${\tan }^{-1}\left|\frac{\left(a-b\right)}{1+ab}\right|$
• B. ${\tan }^{-1}\left|\frac{\left(a-b\right)}{1-ab}\right|$
• C. ${\tan }^{-1}\left|\frac{\left(\log b+\log a\right)}{1+\log a\log b}\right|$
• D. ${\tan }^{-1}\left|\frac{\left(\log b+\log a\right)}{1-\log a\log b}\right|$
• E. ${\tan }^{-1}\left|\frac{\left(\log b-\log a\right)}{1+\log a\log b}\right|$

Answer 1

The correct option is E

${\tan }^{-1}\left|\frac{\left(\log b-\log a\right)}{1+\log a\log b}\right|$

Explanation for the correct option:

Step 1: Finding the slope $m_1$ and $m_2$

Given $y=a^x$ and substituting the value of $y=b^x$

$$\begin{gathered} \Rightarrow b^x=a^x \\ \Rightarrow {\left(\frac{a}{b}\right)}^x=1 \\ \Rightarrow x=0 \end{gathered}$$

Given

$y=a^x$ and $y=b^x$

Differentiating the above equations with respect to $x$

$$\begin{gathered} y=a^x \\ y\text{'}=a^x\log a \\ {m}_1=\log a\mathbf{}\left[\text{At}\left(0,1\right)\right] \end{gathered}$$

And

$$\begin{gathered} y=b^x \\ y\text{'}=b^x\log b \\ {m}_2=\log b\left[\text{At}\left(0,1\right)\right] \end{gathered}$$

Step 2: Finding the angle between the curves

We know that

$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

Substitute the value of $m_1\text{and}{m}_2$ in the above Equation

$$\begin{gathered} \tan \theta =\left|\frac{\log b-\log a}{1+\log a\log b}\right| \\ \theta ={\tan }^{-1}\left|\frac{\log b-\log a}{1+\log a\log b}\right| \end{gathered}$$

Hence, option (E) is the correct answer.