Answer:
[tex]\ln (y) = 0.210x-2.601[/tex]
Step-by-step explanation:
Given table:
[tex]\begin{array}{|c|c|}\cline{1-2} \sf Length & \sf Diameter \\x & y\\\cline{1-2} \vphantom{\dfrac12} 2 & 0.113\\\cline{1-2} \vphantom{\dfrac12} 3& 0.151\\\cline{1-2} \vphantom{\dfrac12} 4& 0.192\\\cline{1-2} \vphantom{\dfrac12} 5& 0.222\\\cline{1-2} \vphantom{\dfrac12} 6& 0.262\\\cline{1-2} \end{array}[/tex]
To convert y = abˣ to linear form, take natural logs of both sides and rearrange:
[tex]\begin{aligned}y=ab^x \implies \ln y &= \ln ab^x\\\implies \ln y &= \ln a + \ln b^x\\\implies \ln y &= \ln a + x \ln b\\\implies \ln y &=x \ln b+ \ln a\end{aligned}[/tex]
This is in the straight-line form y = mx + c.
Substitute the first and last values of x and y from the table into the natural log formula:
- [tex]\ln 0.113 = 2 \ln b+\ln a[/tex]
- [tex]\ln 0.262 = 6 \ln b+\ln a[/tex]
Subtract the first equation from the second equation to eliminate ln(a);
[tex]\implies \ln 0.262 - \ln 0.113 = 6 \ln b - 2 \ln b[/tex]
[tex]\implies \ln 0.262 - \ln 0.113 = 4 \ln b[/tex]
Apply the quotient log law:
[tex]\implies \ln \dfrac{0.262}{0.113} = 4\ln b[/tex]
Rearrange and solve for ln(b):
[tex]\implies \ln b = \dfrac{1}{4}\ln \dfrac{0.262}{0.113}[/tex]
[tex]\implies \ln b = 0.210\; \sf (3\;d.p.)[/tex]
Substitute the found value of ln(b) into one of the equations and solve for ln(a):
[tex]\implies \ln 0.262 = 6 \cdot \dfrac{1}{4}\ln \dfrac{0.262}{0.113}+\ln a[/tex]
[tex]\implies \ln a=\ln 0.262 -\dfrac{6}{4}\ln \dfrac{0.262}{0.113}[/tex]
[tex]\implies \ln a = -2.601\; \sf (3\;d.p.)[/tex]
Substitute the found values of ln(a) and ln(b) into the formula to create an equation for ln(y):
[tex]\boxed{\ln (y) = 0.210x-2.601}[/tex]
