Can one determine $\lambda$ such that $C_1(\gamma^2+\gamma)\gamma^{-2\beta \lambda}=C_2(\gamma+1)\gamma^{(2\alpha+1)\lambda}$? I was wondering if one could determine $\lambda$ such that $C_1(\gamma^2+\gamma)\gamma^{-2\beta \lambda}=C_2(\gamma+1)\gamma^{(2\alpha+1)\lambda}$
This expression is equivalent to $C_1(\gamma^{2-2\beta\lambda}+\gamma^{1-2\beta\lambda})=C_2(\gamma^{1+(2\alpha+1)\lambda}+\gamma^{(2\alpha+1)\lambda})$, but this seems to be impossible to solve for $\lambda$. Am I correct in my hypothesis?
 A: $$C_1(\gamma^2+\gamma)\gamma^{-2\beta \lambda}=C_1(\gamma+1)\gamma \gamma^{-2\beta \lambda}=C_1(\gamma+1)\gamma^{1-2\beta \lambda}=C_2 (\gamma +1) \gamma^{(2 \alpha +1)\lambda}$$
If $\gamma \neq -1$, we can divide by $(\gamma+1)$ both sides (If $\gamma=-1$ or $\gamma=0$, all of the possible $\lambda$ are good; maybe the $0$ is not good if $\gamma=0$, because the $0^0$ is undefined):
$$C_1 \gamma ^{1-2 \beta \lambda}=C_2 \gamma ^{\lambda + 2 \alpha \lambda}$$
$$\frac{C_1}{C_2}=\gamma ^{\lambda (1+2 \alpha +2 \beta)-1}$$
$$\frac{C_1}{C_2} \gamma = \gamma ^ {\lambda (1+2 \alpha + 2 \beta)}$$
$$\log_{\gamma}{\left(\frac{C_1}{C_2} \gamma\right)}=\lambda (1+2 \alpha + 2 \beta)$$
$$\lambda=\frac{\log_{\gamma}{\left(\frac{C_1}{C_2} \gamma\right)}}{1+2 \alpha + 2 \beta}$$
$$\lambda=\frac{1+\log_{\gamma}{\left(\frac{C_1}{C_2}\right)}}{1+2 \alpha + 2 \beta}$$
A: taking the logarithm on both sides we get
$$\ln(C_1(\gamma^2+\gamma)\gamma)-2\beta\lambda\ln(\gamma)=\ln(C_2(\gamma+1)\gamma)+(2\alpha)+1)\lambda\ln(\gamma)$$
rearranging
and we obtain
$$\ln(C_1(\gamma^2+\gamma)\gamma)-\ln(C_2(\gamma+1)\gamma))=\lambda(2\beta\ln(\gamma)+(2\alpha+1)\ln(\gamma))$$ and if
$$2\beta\ln(\gamma)+(2\alpha+1)\ln(\gamma)\neq 0$$ we have
$$\lambda=\frac{\ln(C_1(\gamma^2+\gamma)\gamma)-\ln(C_2(\gamma+1)\gamma))}{2\beta\ln(\gamma)+(2\alpha+1)\ln(\gamma)}$$
