Solution of an exponential equation Probably very simple question. Why the solution of
$$1=n(1-a)^{t}$$
in terms of $t$ is equal to:
$$t=\frac{\ln n}{\ln \frac{1}{1-a}}$$
 A: we can write $\frac{1}{n}=(1-a)^t$ taking the logarithm of both sides we get $$-\ln(n)=t\ln(1-a)$$ and thus we get $$t=-\frac{\ln(n)}{\ln(1-a)}$$
A: Do you know that $\ln 1=0$ and that $$\ln \left(n(1-a)^t\right)=t \ln (n(1-a))=t \left(\ln n +\ln(1-a)\right)$$ and that $$\ln(1-a)=-(-\ln(1-a))=-\ln(1-a)^{-1}=-\ln \frac{1}{1-a}$$ which are due to the properties of the logarithm


*

*$\ln 1=0$,

*$\ln a^x=x\ln a$,

*$\ln ab=\ln a+ \ln b$

A: $$
\ln(1) = 0 = \ln\left(n(1-a)^t\right)
$$
Now the later can be expanded using
$$
\ln(ab) = \ln a + \ln b
$$ 
Thus
$$
\ln\left(n(1-a)^t\right)  = \ln n + \ln \left[(1-a)^t\right]\tag{*}
$$
Using
$$
\ln a^b = b\ln a
$$
We can rewrite Eq * as 
$$
\ln n + t\ln (1-a) = 0
$$
You can rearrange for t now. Now the final bit of the puzzle is using the fact
$$
-\ln(1-a) = \ln\left(\frac{1}{1-a}\right)
$$
A: Hint: $$0= \ln 1 =\ln \Big(n(1-a)^t\Big) \Rightarrow \ln n + t\ln(1-a) = 0 $$
A: Hint: You can write $n(1-a)^t = e^{t\ln (1-a) + \ln n}$
A: $$\ln 1=\ln\left(n\left(1-a\right)^t\right)$$
$$0=\ln n+t\ln(1-a)$$
$$t=-\frac{\ln n}{\ln(1-a)}$$
A: Notice that:
$$\ln 1 = \ln\left(n(a-1)^t\right)$$
$$0 =  \ln\left(n(a-1)^t\right)$$
$$0 = \ln\left(\frac{n}{\frac{1}{(a-1)^t}}\right) = \ln n - \ln\left(\frac{1}{(a-1)^t}\right) = \ln n + t\ln (a-1)$$
$$t = \frac{\ln n }{-\ln (a-1)} = \frac{\ln n}{\ln \frac{1}{a-1}}$$
A: Given $$1=n(1-a)^t \implies (1-a)^{-t}=n \implies \big(\frac {1}{1-a}\big)^{t}=n$$
Now taking logarithm on both sides we get, $$t\ln{\frac{1}{1-a}}=\ln{n} \implies t=\frac{\ln{n}}{\ln{\frac{1}{1-a}}}$$
