Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
Usually $k$ is used for integers. Is that the case here? If so then you only have to check a few. –  drhab Apr 28 at 12:31
    
@Yiyuan I edited –  Peterix Apr 28 at 12:46
    
worth a +1 plz? –  Peterix Apr 28 at 13:20

2 Answers 2

up vote 1 down vote accepted

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. $k=2,3, \dots$

share|improve this answer
    
Thanks I actually never thought about that, it was easier then I expected –  Peterix Apr 28 at 15:19

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.