Best fit line on Semi log graph

Hello, I'm new to matlab. i want to plot a graph where x axis is in log scale and y axis is linear. The graph beow though shows straight line but I know the equation is not correct because It didn't pass throgh the scatter points

here is the code. Please help me, how can these line pass through the points and also how can i get the equation?

data=[0.891  35784.525582
      0.91   39142.72
      0.815  17679.26
      0.891  25582
      0.8861 30168.052]
 
p1=data(:,2); %Re
q1=data(:,1); %Cd
scatter(p1,q1)

set(gca,'XScale','log');

pp = polyfit(log(p1), q1, 1)

qq=exp(polyval(pp,log(p1)))
hold on
plot(p1, qq)
hold off

Answer 1

[Alan Stevens]

Like so

data=[0.891  35784.525582
      0.91   39142.72
      0.815  17679.26
      0.891  25582
      0.8861 30168.052];
 
p1=data(:,2); %Re
q1=data(:,1); %Cd
logp1 = log10(p1);

pp = polyfit(logp1, q1, 1);

qq=polyval(pp,logp1);

plot(logp1, qq,'-*',logp1,q1,'o'),grid
xlabel('log10(p1)'),ylabel('q1')
legend('fit','data')

[Maiesha Nujhat]

thanks, the line goes through the points now.. but can you help me further? in excel I plotted this graph but the equation shows something else. the picture is given below. From the code I get

pp =

    0.2460   -0.2177

can you tell me where is the problem?

[Alan Stevens]

The Matlab figures come from fitting Cd to lo10(Re).

The Excel figures come from fitting log10(Cd) to log10(Re).

If we modify the MATLAB program to the following

data=[0.891  35784.525582
      0.91   39142.72
      0.815  17679.26
      0.891  25582
      0.8861 30168.052];
 
p1=data(:,2); %Re
q1=data(:,1); %Cd
logp1 = log10(p1);
logq1 = log10(q1);
pp = polyfit(logp1, logq1, 1);
qq=polyval(pp,logp1);
plot(logp1, qq,'-*',logp1,logq1,'o'),grid
xlabel('log10(Re)'),ylabel('log(Cd)')
legend('fit','data')

we get

pp =

    0.1244   -0.6110

We can manipulate this as follows

%  pp = 0.1244   -0.6110
%  m = pp(1);  c = pp(2)
%  log10(Cd) = m*log10(Re) + c
%  log10(Cd) = log10(Re^m) + c
%  Cd = (10^c)*Re^m
%        10^c = 0.2449
%  Cd = 0.2449*Re^0.1244

So the result depends on the type of fit, Cd vs log10(Re) or log10(Cd) vs log10(Re).

[Maiesha Nujhat]

Thank you so much!