Answer:
[tex]\ln (y) = -0.250 \ln (x)+0.223[/tex]
Step-by-step explanation:
Given table:
[tex]\begin{array}{|c|c|}\cline{1-2} \vphantom{\dfrac12}x & y\\\cline{1-2} \vphantom{\dfrac12} 1 & 1.250\\\cline{1-2} \vphantom{\dfrac12} 2 & 1.051\\\cline{1-2} \vphantom{\dfrac12} 3& 0.950\\\cline{1-2} \vphantom{\dfrac12} 4& 0.884\\\cline{1-2} \vphantom{\dfrac12} 5& 0.836\\\cline{1-2} \vphantom{\dfrac12} 6& 0.799\\\cline{1-2} \end{array}[/tex]
To convert y = axⁿ to linear form, take natural logs of both sides and rearrange:
[tex]\begin{aligned}y=ax^n \implies \ln y &= \ln ax^n\\\implies \ln y &= \ln a + \ln x^n\\\implies \ln y &= \ln a + n \ln x\\\implies \ln y &=n \ln x+ \ln a\end{aligned}[/tex]
This is in the straight-line form y = mx + c.
Substitute the first values of x and y from the table into the natural log formula and solve for ln(a):
[tex]\implies \ln 1.250= n \ln 1+\ln a[/tex]
[tex]\implies \ln 1.250= n (0)+\ln a[/tex]
[tex]\implies \ln a = \ln 1.250[/tex]
[tex]\implies \ln a = 0.223\; \sf (3 \; d.p.)[/tex]
Substitute the found value of ln(a) and the last values of x and y from the table into the natural log formula and solve for n:
[tex]\implies \ln 0.799= n \ln 6+\ln 1.250[/tex]
[tex]\implies n \ln 6=\ln 0.799-\ln 1.250[/tex]
[tex]\implies n =\dfrac{\ln 0.799-\ln 1.250}{\ln 6}[/tex]
[tex]\implies n =-0.250\;\sf (3 \; d.p.)[/tex]
Substitute the found values of n and ln(a) into the formula to create an equation for ln(y):
[tex]\boxed{\ln (y) = -0.250 \ln (x)+0.223}[/tex]
