solve three non linear power equations

I have three equations. I want to solve kindly help me.

x*10^y+z=5
x*25^y+z=4.95
x*30^y+z=4.89

Answer 1

[Star Strider]

Try this (requires the Symbolic Math Toolbox):

syms x y z 
assume(x > 0)
M = [x*10^y+z==5;  x*25^y+z==4.95;  x*30^y+z==4.89];
[xsol,ysol zsol] = vpasolve(M, [x,y,z], 'random',true)

There are several solutions. Experiment with starting values and different states for 'random'.

Answer 2

[Sarah Gilmore]

To solve your system of three non-linear equations, you can use the fsolve function. In one matlab function, define the following:

function F = root3d(x)
  F(1) = x(1)*10^x(2) + x(3) - 5;
  F(2) = x(1)*25^x(2) + x(3) - 4.95;
  F(3) = x(1)*30^x(2) + x(3) - 4.89;
end

Note that the three equations are written in F(x) = 0 form and the three variables, x, y and z are replaced by x(1), x(2), x(3). x is a 1x3 vector in this case. Now, in a separate matlab script file write the following code:

x0 = [5 4.95 4.89];
x = fsolve(@root3d, x0)

x0 is your initial guess for the values of x, y and z fsolve uses to the solve the system. The first argument in fsolve, @root3d, is an anonymous function. The second argument, x0, is the input passed to root3d.

To learn more about anonymous functions, check out the following link to matlab's documentation:

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

To learn more about fsolve and look at more examples, check out this link:

https://www.mathworks.com/help/optim/ug/fsolve.html#inputarg_x0

Answer 3

[John D'Errico]

Do it mainly with pencil and paper?

x*(10^y - 25^y) = 5 - 4.95
x*(10^y - 30^y) = 5 - 4.89

Now take the ratio.

(10^y - 25^y)/(10^y - 30^y) = 0.05/0.11

Some simple algebra.

(.06/0.11)*10^y - 25^y + (0.05/0.11)*30^y == 0

You could use fzero to solve this, or use the symbolic toolbox. Plot shows two solutions.

vpasolve((.06/0.11)*10^y - 25^y + (0.05/0.11)*30^y == 0,4)
ans =
4.264083495647992647900412924213

vpasolve((.06/0.11)*10^y - 25^y + (0.05/0.11)*30^y == 0)
ans =
0

Is the solution at zero a valid solution? Perhaps not, since that would result in a divide by zero. In fact, the original equations make y==0 obviously not a solution. You can go back and recover x and z now.

Or, better yet, divide by 10^y, which is fine, since 10^y will never be zero.

(.06/0.11) - 2.5^y + (0.05/0.11)*3^y == 0

Again, we can see that y==0 is a trivial solution, but not admissible, since that solution was generated by algebraic operations we did before.

It is homework though, so your problem to solve.