Solution of an exponential equation

Question

Probably very simple question. Why the solution of

$$1=n(1-a)^{t}$$

in terms of $t$ is equal to:

$$t=\frac{\ln n}{\ln \frac{1}{1-a}}$$

Answer 1

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 $$t=-\frac{\ln(n)}{\ln(1-a)}$$

Answer 2

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

1. $\ln 1=0$,
2. $\ln a^x=x\ln a$,
3. $\ln ab=\ln a+ \ln b$

Answer 3

$$ \ln(1) = 0 = \ln\left(n(1-a)^t\right) $$ 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 $$ -\ln(1-a) = \ln\left(\frac{1}{1-a}\right) $$

Answer 4

Hint: $$0= \ln 1 =\ln \Big(n(1-a)^t\Big) \Rightarrow \ln n + t\ln(1-a) = 0 $$

Answer 5

Hint: You can write $n(1-a)^t = e^{t\ln (1-a) + \ln n}$

Answer 6

$$\ln 1=\ln\left(n\left(1-a\right)^t\right)$$

$$0=\ln n+t\ln(1-a)$$

$$t=-\frac{\ln n}{\ln(1-a)}$$

Answer 7

Notice that:

$$\ln 1 = \ln\left(n(a-1)^t\right)$$

$$0 = \ln\left(n(a-1)^t\right)$$

$$0 = \ln\left(\frac{n}{\frac{1}{(a-1)^t}}\right) = \ln n - \ln\left(\frac{1}{(a-1)^t}\right) = \ln n + t\ln (a-1)$$

$$t = \frac{\ln n }{-\ln (a-1)} = \frac{\ln n}{\ln \frac{1}{a-1}}$$

Answer 8

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, $$t\ln{\frac{1}{1-a}}=\ln{n} \implies t=\frac{\ln{n}}{\ln{\frac{1}{1-a}}}$$