Exponential function to logarithmic function

i'm stuck on completing this equations. Is this correct?

$$z=a e^{-bt}$$

$$\ln(z)=\ln(a)+\ln(e^{-bt})$$

$$\ln(z)=\ln(a)+(1)(-bt)$$

$$\ln(z)=\ln(a)-bt$$

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Yes, it is. Using the rules of logarithms: $$ln(z)=ln(a)+ln(e^{-bt})$$ $$ln(z)=ln(a)-bt \cdot [ln(e)]$$ $$ln(z) = ln(a)-bt$$ Which is exactly what you made. – A. A. Jul 17 '14 at 15:46
You want "exponential function" and "logarithmic function" instead of "polynomial function" and "linear function" here. – Antonio Vargas Jul 17 '14 at 15:53

It is correct!!!

If you want to solve for $t$, it is as followed:

$$\ln{(z)}=\ln{(a)}-bt \Rightarrow bt=\ln{(a)}-\ln{(z)}=\ln{(\frac{a}{z})} \Rightarrow t=\frac{1}{b} \ln{(\frac{a}{z})}, \ \text{ where } z,a>0$$

Like @Alex R. says in the comment, from the relation $$z=a e^{-bt}$$ if $\displaystyle{z=0 \Rightarrow a=0 \text{ and } b,t \text{ are arbitrary }.}$

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And if $z=0$ it implies $a=0$ and $b,t$ arbitrary. – AlexR Jul 17 '14 at 16:13
Yes, you did apply well this property of logarithm: $ln(ab)=ln(a) + ln(b)$, and the fact that $ln(x)$ is the inverse of $e^x$