Solving sets of 3 equations with 3 unknown with some logarithms thrown in.

Question

I'm having some difficulty solving the following set of equations:

$6.16=x\log{(100y+z)}\\ 8.59=x\log{(250y+z)}\\ 12.72=x\log{(1000y+z)}$

The sum of the two different variables in the logarithm is throwing me off and I don't know how to solve it.

Answer 1

I don't think there's a nice solution, but I did this:

Let $a=1/x$, $b=50y$, and $c=100y+z.$ Then the equations become

$$6.16a=\log c$$

$$8.59a=\log (c+3b)$$

$$12.72a =\log (c+18b)$$

Exponentiate:

$$\exp(6.16a) = c$$

$$\exp(8.59a) = c+3b$$

$$\exp(12.72a) = c+18b$$

Eliminate $c$:

$$\exp(8.59a) = \exp(6.16a)+3b$$

$$\exp(12.72a) = \exp(6.16a)+18b$$

Take $6$ times the frist equation and subtract the second to get rid of $b$:

$$6\exp(8.59a) -\exp(12.72a) = 5\exp(6.16a)$$

Divide through by $\exp(6.16a)$:

$$6\exp(2.43a) - \exp(6.56a) = 5.$$

Now get a numerical solution by your favorite method. Maple gives me $a=0$ (which won't work) and $a=0.3151672618.$

So $x = 1/a = 3.172918387.$