Can someone solve this inequality for $T$

$$ R\gamma^T(1/(1−\gamma))≤\epsilon $$ In a paper it solved for $T$ and the inequation below is the result, but I can not prove how the inequation above can be the inequation below: $$ T≥(1/(1−\gamma))\log⁡(R/(\epsilon(1−\gamma))) $$ May someone help me,

thanks in advanced.

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    $\begingroup$ What are $\;\gamma, R, \epsilon...\;$ ? Real numbers, positive, perhaps $\;0<\gamma<1\;$ ? $\endgroup$ – Timbuc Dec 24 '14 at 16:42
  • $\begingroup$ All are real numbers, and $0<\epsilon,\gamma<1$! Thancks for your comment @Timbuc $\endgroup$ – S0H31L Dec 24 '14 at 16:43
  • $\begingroup$ @So Are you asking me?! $\endgroup$ – Timbuc Dec 24 '14 at 16:44
  • $\begingroup$ Sorry! I edited my previous comment $\endgroup$ – S0H31L Dec 24 '14 at 16:45
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    $\begingroup$ I think also $\;R>0\;$ , otherwise the inequality is trivially true for all $\;t\in\Bbb R\;$ . $\endgroup$ – Timbuc Dec 24 '14 at 16:47

We assume that $0<\gamma<1$, $R>0$, $\epsilon>0$. We multiply both sides of $$ R\gamma^T(1/(1−\gamma))≤\epsilon $$ by the positive real number $\displaystyle (1−\gamma)/R$ to get $$ \gamma^T \leq\epsilon(1−\gamma)/R $$ since $\displaystyle (\infty,0) \ni x \mapsto \log x$ is an increasing function we get $$ T \log \gamma \leq \log \left(\epsilon(1−\gamma)/R \right) $$ equivalently $$ -T \log \gamma \geq \log \left(R/(\epsilon(1−\gamma)) \right) $$ but how to get the desired inequality since we know that $1-\gamma \geq \log \gamma$, $0<\gamma<1$.


Assuming everything has the correct value (i.e., all is positive and $\;0<\gamma <1\;$):

$$R\gamma^t\frac1{1-\gamma}\le\epsilon\implies \gamma^t\le\frac{\epsilon(1-\gamma)}R\implies t\log\gamma\le\log\frac{\epsilon(1-\gamma)}R\implies$$


I've no idea how in that paper they got the inequality you give for $\;t\;$ .

  • $\begingroup$ I think using the assumption in @Olivier's answer we can replace $log\gamma$ with $1-\gamma$, may you confirm it? $\endgroup$ – S0H31L Dec 24 '14 at 17:18
  • $\begingroup$ @S0H31L I can't see how could you do that, but it is not only that: the other logarithm's argument is upside down. $\endgroup$ – Timbuc Dec 24 '14 at 18:29
  • $\begingroup$ oh no! you are right! :s tnx! $\endgroup$ – S0H31L Dec 24 '14 at 18:43

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