Use the linear approximation method to estimate the value of ln (2.7^2) in terms of e
G Guest Guest Oct 31, 2005 #1 Use the linear approximation method to estimate the value of ln (2.7^2) in terms of e
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Oct 31, 2005 #2 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 ln(2.72)\displaystyle \ln(2.7^2)ln(2.72) in terms of e\displaystyle ee Click to expand... Let y = ln(x2) = 2 ln(x)\displaystyle y \:= \:\ln(x^2) \:= \:2\,\ln(x)y=ln(x2)=2ln(x) Then dy = 2 dxx\displaystyle dy \,= \,\frac{2\,dx}{x}dy=x2dx Since 2.7 ≈ e − 0.0183\displaystyle 2.7\ \approx\ e\,-\,0.01832.7 ≈ e−0.0183, we will use: x=1, dx = −0.0183\displaystyle x = 1,\;dx\,=\,-0.0183x=1,dx=−0.0183 . . Then: .dy = 2(−0.0183)e = −0.01346388\displaystyle dy\ = \ \frac{2(-0.0183)}{e}\ =\ -0.01346388dy = e2(−0.0183) = −0.01346388 Hence: .2 ln(e − 0.0183) ≈ y + dy = 2 ln(e) − 0.01346388\displaystyle 2\,\ln(e\,-\,0.0183)\ \approx\ y\,+\,dy\ = \ 2\,\ln(e)\,-\,0.013463882ln(e−0.0183) ≈ y+dy = 2ln(e)−0.01346388 Therefore: .ln(2.72) ≈ 1.9865⏟35612\displaystyle \ln(2.7^2)\ \approx\ \underbrace{1.9865}35612ln(2.72) ≈ 1.986535612 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ This approximation is very good . . . The actual value is: .ln(2.72) = ln(7.29) = 1.9865⏟03546\displaystyle \ln(2.7^2) \ =\ \ln(7.29)\ = \ \underbrace{1.9865}03546ln(2.72) = ln(7.29) = 1.986503546 Given: ln(7.29) = 1.9865, then e1.9865 = 7.28997...\displaystyle \text{Given: }\ln(7.29)\ =\ 1.9865\text{, then }e^{1.9865}\ =\ 7.28997...Given: ln(7.29) = 1.9865, then e1.9865 = 7.28997... . . . . close enough!
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 ln(2.72)\displaystyle \ln(2.7^2)ln(2.72) in terms of e\displaystyle ee Click to expand... Let y = ln(x2) = 2 ln(x)\displaystyle y \:= \:\ln(x^2) \:= \:2\,\ln(x)y=ln(x2)=2ln(x) Then dy = 2 dxx\displaystyle dy \,= \,\frac{2\,dx}{x}dy=x2dx Since 2.7 ≈ e − 0.0183\displaystyle 2.7\ \approx\ e\,-\,0.01832.7 ≈ e−0.0183, we will use: x=1, dx = −0.0183\displaystyle x = 1,\;dx\,=\,-0.0183x=1,dx=−0.0183 . . Then: .dy = 2(−0.0183)e = −0.01346388\displaystyle dy\ = \ \frac{2(-0.0183)}{e}\ =\ -0.01346388dy = e2(−0.0183) = −0.01346388 Hence: .2 ln(e − 0.0183) ≈ y + dy = 2 ln(e) − 0.01346388\displaystyle 2\,\ln(e\,-\,0.0183)\ \approx\ y\,+\,dy\ = \ 2\,\ln(e)\,-\,0.013463882ln(e−0.0183) ≈ y+dy = 2ln(e)−0.01346388 Therefore: .ln(2.72) ≈ 1.9865⏟35612\displaystyle \ln(2.7^2)\ \approx\ \underbrace{1.9865}35612ln(2.72) ≈ 1.986535612 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ This approximation is very good . . . The actual value is: .ln(2.72) = ln(7.29) = 1.9865⏟03546\displaystyle \ln(2.7^2) \ =\ \ln(7.29)\ = \ \underbrace{1.9865}03546ln(2.72) = ln(7.29) = 1.986503546 Given: ln(7.29) = 1.9865, then e1.9865 = 7.28997...\displaystyle \text{Given: }\ln(7.29)\ =\ 1.9865\text{, then }e^{1.9865}\ =\ 7.28997...Given: ln(7.29) = 1.9865, then e1.9865 = 7.28997... . . . . close enough!
