Solving an equation with log

Hi i'm fairly new to MATLAB and encounter a problem regarding this equation y=c(x)^m

where m is the gradient of points:

(x,1y1)=(100,50)

(x2,y2)=(1000,10)

This is the eq i put on MATLAB:

Eq = log10(Y2) == log10(C1*X2^(m)); %Equation
C1 = vpasolve (Eq, C1)

It seems that i get C far from my hand-drawn answer

How to solve C?

Answer 1

[Walter Roberson]

Looks okay to me.

format long g
X1 = 100; Y1 = 50;
X2 = 1000; Y2 = 10;
m = (Y2-Y1)./(X2-X1);
syms C1
Eq = log10(Y2) == log10(C1*X2^(m)); %Equation
C1sol = solve(Eq,C1)

C1sol = 

vpa(C1sol)

ans = 

13.593563908785257310765717430783

%log10(Y2) == log10(C1*X2^m) implies
%Y2 == C1*X2^m implies
C1_numeric = Y2/(X2^m)

C1_numeric = 
          13.5935639087853

[Rafi B]

Hi Walter, thank you so much for the feedback.

The answer for C given by the book is still way off, I'm not sure if it is wrong

or i misinterpret the question.

Any thoughts?

[Walter Roberson]

syms x1 x2 y1 y2
syms c m
eqn1 = y1 == c*x1^m
eqn2 = y2 == c*x2^m
eqn3 = y1/y2 == (c*x1^m)/(c*x2^m)

You should be able to proceed from here, as the above eliminates one of the variables, so re-arrange to solve the other variable.

[Rafi B]

quick question, on eqn3 wouldn't it just cross the c value off?

[Walter Roberson]

Yes, giving you an equation of the form A=B^m with known A and B, which you can use to find m easily.