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$$162\left(1-\left(\frac{1}{3}\right)^n\right) -162=-0.05$$ 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

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    $\begingroup$ The right side should possibly say = -0.05 and then 7.36 would work for n $\endgroup$ – Paul Apr 10 '15 at 11:15
  • $\begingroup$ It is definitely so, the right side should be $-0.05$. $\endgroup$ – Andreas Caranti Apr 10 '15 at 11:22
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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 ? $$162(1-(\frac{1}{3})^n) -162=+0.05$$

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  • $\begingroup$ what is $\frac{1}{3}^n$? $\endgroup$ – Dr. Sonnhard Graubner Apr 10 '15 at 11:24
  • $\begingroup$ I mean $(\frac{1}{3})^n$ $\endgroup$ – Khosrotash Apr 10 '15 at 11:38
  • $\begingroup$ Thanks lads but solved it - book is wrong, doesn't solve for + 0.05 as it should be -0.05 $\endgroup$ – Paul Apr 10 '15 at 11:50
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Here are the steps $$ 162\left(1-\left(\frac13\right)^n\right) -162=-0.05 $$ $$ 162\left(1-\left(\frac13\right)^n -1\right)=-0.05 $$ $$ 162\left(-\left(\frac13\right)^n \right)=-0.05 $$ $$ -162\left(\frac13\right)^n =-0.05 $$ $$ 162\left(\frac13\right)^n =0.05 $$ $$ \left(\frac13\right)^n =\frac{0.05}{162}$$ $$\ln \left(\frac13\right)^n =\ln\left(\frac{0.05}{162}\right)$$ $$ n\ln \left(\frac13\right) =\ln\left(\frac{0.05}{162}\right)$$ $$ n =\frac{\ln\left(\frac{0.05}{162}\right)}{\ln \left(\frac13\right)} =-\frac{\ln\left(\frac{0.05}{162}\right)}{\ln \left(3\right)} $$ $$= \frac{\ln\left(162\right) -\ln\left(0.05\right)}{\ln \left(3\right)}= \frac{\ln\left(2\cdot 3^4\right) -\ln\left(\frac1{20}\right)}{\ln \left(3\right)} $$ $$= \frac{\ln\left(2\right)+4\ln\left(3\right) +\ln\left(20\right)}{\ln \left(3\right)} = \frac{\ln\left(2\right)+4\ln\left(3\right) +\ln\left(2^2\cdot 5\right)}{\ln \left(3\right)} $$ $$= \frac{\ln\left(2\right)+4\ln\left(3\right) +2\ln\left(2\right)+\ln\left(5\right)}{\ln \left(3\right)}= \frac{3\ln\left(2\right) +\ln\left(5\right)}{\ln \left(3\right)} +4$$

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