Hi guys I am trying to do Lagrange multipliers to figure out $\lambda$ $$F=a \log(x^2-y)+b\log(x^3-z)-\lambda (x^2-y+x^3-z -1)$$ Where a and b are constants and we have the constraint $x^2-y+x^3-z =1$
What I did was take the partials and added them together and got $a(\frac{2x-1}{x^2-y})+b(\frac{3x-1}{x^3-z})- \lambda (2x+3x^2)+2 \lambda$
I am not sure how to find this lambda
Suppose your $\log$ means natural logarithm. Then set the partial derivatives to $0$: $$\frac{2ax}{x^2-y}+\frac{3bx^2}{x^3-z}-2x\lambda+3x^2\lambda=0 \\ \frac{-a}{x^2-y}+\lambda=0 \\ \frac{-b}{x^3-z}+\lambda=0\\ x^2-y+x^3-z =1$$ From equation 2 and 3: $$\lambda=\frac{a}{x^2-y}=\frac{b}{x^3-z}$$ Plug these information into 1: $$2x\lambda+3x^2\lambda-2x\lambda+3x^2\lambda=0$$
Can you go on from here?