Question about $z=-e^{\ln(n)\cdot(-x-2)+\ln(m)\cdot(x-1)}$

Question

If we know that (where all the variables are real and positive):

$$z=-e^{\ln(n)\cdot(-x-2)+\ln(m)\cdot(x-1)}$$

What will this be:

$$e^{\ln(n)\cdot(-x-2)+\ln(m)\cdot(x+1)}$$

in terms of $z$?

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My work:

I can write:

$$z=-e^{\ln(n)\cdot(-x-2)+\ln(m)\cdot(x-1)}=-e^{\ln(n)\cdot(-x-2)}\cdot e^{\ln(m)\cdot(x-1)}$$

And then I do not now how to proceed.

Answer 1

Using $e^{x \ln a} = e^{\ln a^{x}} = a^{x}$ then $$-z = e^{-(x+2) \ln(n) + (x-1) \ln m} = \frac{1}{m \, n^{2}} \, \left(\frac{m}{n}\right)^{x}.$$ Now \begin{align} e^{-(x+2) \ln(n) + (x+1) \ln m} &= \frac{m}{n^{2}} \, \left(\frac{m}{n}\right)^{x} \\ &= m^{2} \cdot \frac{1}{m \, n^{2}} \, \left(\frac{m}{n}\right)^{x} \\ &= - m^{2} \, z. \end{align}

This can also be seen as: \begin{align} e^{-(x+2) \ln(n) + (x+1) \ln m} &= e^{-(x+2) \ln(n) + (x-1) \ln m + 2 \ln m} \\ &= m^{2} \, e^{-(x+2) \ln(n) + (x-1) \ln m} \\ &= - m^{2} \, z \end{align}

Answer 2

Hint

Call:

$$g=e^{\ln\left(\text{n}\right)\cdot\left(-x-2\right)+\ln\left(\text{m}\right)\cdot\left(x+1\right)}$$

and calculate:

$$\frac{g}{z}=-\frac{e^{\ln\left(\text{n}\right)\cdot\left(-x-2\right)+\ln\left(\text{m}\right)\cdot\left(x-1\right)}}{e^{\ln\left(\text{n}\right)\cdot\left(-x-2\right)+\ln\left(\text{m}\right)\cdot\left(x+1\right)}}$$

Ps.: Remmember that $$\frac{e^a}{e^b}=e^{a-b}$$

Answer 3

Maybe you can use the fact that $a\log(b)=\log(b^{a})$ and $e^{\log(b)}=b$.