# How I can solve this inequality $(R\gamma^T(1/(1−\gamma))≤\epsilon)$ for $T$?!

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,

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

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

$$t\ge\frac{\log\frac{\epsilon(1-\gamma)}R}{\log\gamma}\;,\;\;\text{since}\;\;\log\gamma<0$$

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

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