Q. 75 - Calculus

Question

Prove that $\ln \left(ab\right)=\ln a+\ln b$ for any $a,b>0$ by following these steps:

(a) Use Theorem 4.35 to show that $\frac{d}{dx}\left(\ln ax\right)=\frac{1}{x}$ for any real number $a>0$.

(b) Use your answer to part (a) to argue that $\ln ax=\ln x+C$. (Hint: Think about Theorem 4.14.)

(c) Solve for the constant $C$ in the equation from the previous part, by evaluating the equation at $x=1$. Use your answer to show that $\ln ax=\ln x+\ln a$. Why does this argument complete the proof?

Step-by-Step Solution

Answer

Part (a) $\frac{d}{dx}\left(\ln ax\right)=\frac{1}{x}$ for any real number $a>0$.

Part (b) $\ln ax=\ln x+C$.

Part (c) $\ln ax=\ln x+\ln a$.

Step 1. Given information

We have to prove that $\ln \left(ab\right)=\ln a+\ln b$ for any $a,b>0$.

Part (a) Step 1. Prove that d d x ln   a x = 1 x for any real number a > 0 .

$$\begin{gathered} \frac{d}{dx}\left(\ln ax\right) \\ =\frac{d}{dx}{\int }_1^{ax}\frac{1}{t}dt \\ =\frac{1}{ax}a \\ =\frac{1}{x} \end{gathered}$$

Part (b) Step 1. Prove that ln   a x = ln   x + C .

$$\begin{gathered} \ln ax \\ =\ln x+\ln a (\ln a is cons\tan t). \\ =\ln x+C \end{gathered}$$

Part (c) Step 1. Solve the constant C .

$$\begin{gathered} \ln ax=\ln x+C \\ \ln a=C+0 \left[x=1,\ln 1=0\right] \\ \ln a=C \end{gathered}$$

Putting $x=b$,

$\ln \left(ab\right)=\ln a+\ln b$