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The question here involved you needing to find the smallest integer $ n $ such that $S_n > 72$ where $$S_N = 6\sum_{i=1}^n 0.92^{i-1} $$ This seemed like a basic question with nothing unusual happening at first but as I started simplifying after using $a_0\frac{(1-r^n)}{1-r} > 72$ I eventually reached $$ 0.92^n < 0.04 $$ and here is when things started getting weird. I took $\log_{0.92}$ on both sides to get $$ n < \log_{0.92} (0.04) $$ Huh? There's no negative sign involved here as far as I can see and since the question is asking for the minimum value $n$ can be it seemed nonsensical to have an inequality showing the upper bound of $n$ rather than the lower bound. Simply because my intuition told me to, I flipped the inequality sign around and ended up getting the question right on my exam but I don't understand why, mathematically, I was supposed to flip the sign. The official answer (as shown in the photo above) used $\log_{10}$ and since $\log_{10} (0.04)$ and $\log_{10}(0.92)$ are both negative it would make sense to flip the sign if you use $\log_{10}$. The guidance for the answer scheme allowed any answers that used $\log_{0.92}$ but only as long as they flipped the inequality sign in that step, without providing any justification. So my question is, why did the inequality sign need to get flipped when I used $\log_{0.92}$ even tho $\log_{0.92}(0.04)$ is positive? According to my limited high school understanding of inequalities, they get flipped only when both sides are multiplied by a negative number and not in any other scenario.

Any help is appreciated

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  • $\begingroup$ Logarithm function with a base $<1$ are very seldom used... $\endgroup$ – Jean Marie Oct 22 '19 at 14:11
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If $0<a<1$, then the function $\log_a$ is a decreasing function. Therefore$$a^n<b\iff\log_a(a^n)>\log_a(b)\iff n>\log_a(b).$$

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  • $\begingroup$ Is it the same procedure for all decreasing functions? $\endgroup$ – Jibran Oct 22 '19 at 17:57
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    $\begingroup$ Yes, by the definition of “decreasing function”. $\endgroup$ – José Carlos Santos Oct 22 '19 at 18:00
  • $\begingroup$ Ah, I've just now come to understand this answer, thanks $\endgroup$ – Jibran Oct 22 '19 at 18:12
  • $\begingroup$ If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ – José Carlos Santos Oct 22 '19 at 18:14
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Since $$\log(0.92)<0$$ you must write $$n>\frac{\log0.04}{\log0.92}$$

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