Algebra and solving for n $$162\left(1-\left(\frac{1}{3}\right)^n\right) -162=-0.05$$
Solve for n
I've tried myself but am getting 2.something and the answer should be 7.36. I know you need to use logs but not working for me 
 A: It is not possible to solve it for natural $n$ , because $$(\frac{1}{3})^n >0\\1-(\frac{1}{3})^n <1\\162(1-(\frac{1}{3})^n) <162\\162(1-(\frac{1}{3})^n) -162<0\\$$and how can be ? $$162(1-(\frac{1}{3})^n) -162=+0.05$$
A: Here are the steps
$$ 162\left(1-\left(\frac13\right)^n\right) -162=-0.05 $$
$$ 162\left(1-\left(\frac13\right)^n -1\right)=-0.05 $$
$$ 162\left(-\left(\frac13\right)^n \right)=-0.05 $$
$$ -162\left(\frac13\right)^n =-0.05 $$
$$ 162\left(\frac13\right)^n =0.05 $$
$$ \left(\frac13\right)^n =\frac{0.05}{162}$$
$$\ln \left(\frac13\right)^n =\ln\left(\frac{0.05}{162}\right)$$
$$ n\ln \left(\frac13\right) =\ln\left(\frac{0.05}{162}\right)$$
$$ n =\frac{\ln\left(\frac{0.05}{162}\right)}{\ln \left(\frac13\right)} =-\frac{\ln\left(\frac{0.05}{162}\right)}{\ln \left(3\right)} $$
$$= \frac{\ln\left(162\right) -\ln\left(0.05\right)}{\ln \left(3\right)}= \frac{\ln\left(2\cdot 3^4\right) -\ln\left(\frac1{20}\right)}{\ln \left(3\right)} $$
$$= \frac{\ln\left(2\right)+4\ln\left(3\right) +\ln\left(20\right)}{\ln \left(3\right)} = \frac{\ln\left(2\right)+4\ln\left(3\right) +\ln\left(2^2\cdot 5\right)}{\ln \left(3\right)} $$
$$= \frac{\ln\left(2\right)+4\ln\left(3\right) +2\ln\left(2\right)+\ln\left(5\right)}{\ln \left(3\right)}= \frac{3\ln\left(2\right) +\ln\left(5\right)}{\ln \left(3\right)} +4$$
