Hello, bobby77!
I haven't done this for a while . . . hope I get it right.
Use the linear approximation method to estimate the value of \(\displaystyle \ln(2.7^2)\) in terms of \(\displaystyle e\)
Let \(\displaystyle y \:= \:\ln(x^2) \:= \:2\,\ln(x)\)
Then \(\displaystyle dy \,= \,\frac{2\,dx}{x}\)
Since \(\displaystyle 2.7\ \approx\ e\,-\,0.0183\), we will use: \(\displaystyle x = 1,\;dx\,=\,-0.0183\)
. . Then:
.\(\displaystyle dy\ = \ \frac{2(-0.0183)}{e}\ =\ -0.01346388\)
Hence:
.\(\displaystyle 2\,\ln(e\,-\,0.0183)\ \approx\ y\,+\,dy\ = \ 2\,\ln(e)\,-\,0.01346388\)
Therefore:
.\(\displaystyle \ln(2.7^2)\ \approx\ \underbrace{1.9865}35612\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This approximation is very good . . .
The actual value is:
.\(\displaystyle \ln(2.7^2) \ =\ \ln(7.29)\ = \ \underbrace{1.9865}03546\)
\(\displaystyle \text{Given: }\ln(7.29)\ =\ 1.9865\text{, then }e^{1.9865}\ =\ 7.28997...\)
. . . . close enough!