So I was reading a math book and I faced with expression I could not solve. Well, I even do not know how to begin, really. I understand that in order to compute power we need to find a logarithm. But... this problem mess everything in my head. Here is the expression and its solution. I would be really happy for detailed steps of how this computation is done. Thank you!

$$1-\left(\dfrac{35}{36} \right)^n \geq \dfrac{1}{2}$$ $$n \geq \dfrac {\ln (\dfrac{1}{2})}{\ln (\dfrac{35}{36})} \approx 24.6$$

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – user109879 Oct 19 '14 at 23:23

$$1 - (\frac{35}{36})^n \geq \frac{1}{2} \Longleftrightarrow - (\frac{35}{36})^n \geq -\frac{1}{2} \Longleftrightarrow \frac{1}{2} \geq (\frac{35}{36})^n \Longleftrightarrow \ln{\frac{1}{2}} \geq \ln {(\frac{35}{36})^n}$$

$$ \ln{\frac{1}{2}} \geq n\ln {(\frac{35}{36})} $$

Now, $\ln {(\frac{35}{36})}$ is negative, so dividing it out and flipping the inequality due to dividing by a negative, we have the desired result.

Note that I used the identity $\ln a^b = b\ln a$.

  • $\begingroup$ what about "$\Longleftarrow$"? $\endgroup$ – anderstood Oct 19 '14 at 23:49
  • $\begingroup$ @anderstood That would also be valid. $\endgroup$ – user109879 Oct 19 '14 at 23:50
  • $\begingroup$ Yes but as it is, you don't prove that the first and last expressions are equivalent, and I don't see why you just considered the direct case "$\Longrightarrow$. $\endgroup$ – anderstood Oct 19 '14 at 23:56
  • $\begingroup$ I interpreted the first as given in the question and the $\implies$ was used merely for simplifying formatting. I can use a double arrow if it will make you happy. $\endgroup$ – user109879 Oct 19 '14 at 23:59
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    $\begingroup$ No pb. I'll try to be less ambiguous too. I just wanted to draw attention on the importance of $\Longleftrightarrow$, $\Longrightarrow$ and $\Longleftarrow$! ;) $\endgroup$ – anderstood Oct 20 '14 at 0:31

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