# 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

• The right side should possibly say = -0.05 and then 7.36 would work for n – Paul Apr 10 '15 at 11:15
• It is definitely so, the right side should be $-0.05$. – Andreas Caranti Apr 10 '15 at 11:22

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 ? $$162(1-(\frac{1}{3})^n) -162=+0.05$$
• what is $\frac{1}{3}^n$? – Dr. Sonnhard Graubner Apr 10 '15 at 11:24
• I mean $(\frac{1}{3})^n$ – Khosrotash Apr 10 '15 at 11:38
Here are the steps $$162\left(1-\left(\frac13\right)^n\right) -162=-0.05$$ $$162\left(1-\left(\frac13\right)^n -1\right)=-0.05$$ $$162\left(-\left(\frac13\right)^n \right)=-0.05$$ $$-162\left(\frac13\right)^n =-0.05$$ $$162\left(\frac13\right)^n =0.05$$ $$\left(\frac13\right)^n =\frac{0.05}{162}$$ $$\ln \left(\frac13\right)^n =\ln\left(\frac{0.05}{162}\right)$$ $$n\ln \left(\frac13\right) =\ln\left(\frac{0.05}{162}\right)$$ $$n =\frac{\ln\left(\frac{0.05}{162}\right)}{\ln \left(\frac13\right)} =-\frac{\ln\left(\frac{0.05}{162}\right)}{\ln \left(3\right)}$$ $$= \frac{\ln\left(162\right) -\ln\left(0.05\right)}{\ln \left(3\right)}= \frac{\ln\left(2\cdot 3^4\right) -\ln\left(\frac1{20}\right)}{\ln \left(3\right)}$$ $$= \frac{\ln\left(2\right)+4\ln\left(3\right) +\ln\left(20\right)}{\ln \left(3\right)} = \frac{\ln\left(2\right)+4\ln\left(3\right) +\ln\left(2^2\cdot 5\right)}{\ln \left(3\right)}$$ $$= \frac{\ln\left(2\right)+4\ln\left(3\right) +2\ln\left(2\right)+\ln\left(5\right)}{\ln \left(3\right)}= \frac{3\ln\left(2\right) +\ln\left(5\right)}{\ln \left(3\right)} +4$$