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I'm studying logarithms and I encountered this equation: $$[\log_9(k+1)]^2+\log_9(k+1)+(k+1)>3$$ I tried a lot but I still couldn't solve it! I know this may be easy for most of you but please could you help me?

Thanks!

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  • $\begingroup$ Usually $k$ is used for integers. Is that the case here? If so then you only have to check a few. $\endgroup$ – drhab Apr 28 '14 at 12:31
  • $\begingroup$ @Yiyuan I edited $\endgroup$ – Peterix Apr 28 '14 at 12:46
  • $\begingroup$ worth a +1 plz? $\endgroup$ – Peterix Apr 28 '14 at 13:20
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With $k \in \mathbb{N}$ you have $\log_9(k+1) > 0.$ Therefore it is obvious that all $k>1$ satisfy the in equality (because $k+1 \ge 3$). For $k=0$ it is obviously wrong, so the only case to check is $k=1$, and here the LHS is $\approx 2.41\dots$.

Summary: The integer solutions are all $k>1$, i.e. $k=2,3, \dots$

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  • $\begingroup$ Thanks I actually never thought about that, it was easier then I expected $\endgroup$ – Peterix Apr 28 '14 at 15:19
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I didn't see if the solution must be integer or real. In every case, pose $t=\log_9(k+1)$, then solve $t^2+t+(k-2)>0$, from which $$ t_{\pm}=\frac{-1\pm\sqrt{1-4(k-2)}}{2} \;.$$ So in order to have real solutions we must impose $k\leq2+\frac{1}{4}$.

Then we have $$ \left[\log_9(k+1)<\frac{-1-\sqrt{1-4(k-2)}}{2}\right]\vee\left[\log_9(k+1)>\frac{-1+\sqrt{1-4(k-2)}}{2}\right]\;. $$

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