Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I try to solve the following equation:

$$ (N+1)^{\log_N{125}} = 216 $$

I know the answer is 5 here but how could I rewrite the equations so I can solve it?

I tried to take the log of both sides but that didn't help me because I got stuck. Could anyone please explain me how to do this?


share|cite|improve this question
up vote 7 down vote accepted

Hint: Use the rules of logarithm, especially the power rule and change of base rule.

Take $\log_6$ on both sides, and then simplify the equation to obtain $\log_6 5 = \log_{N+1} N$.

Observe that the graph of $\log_{N+1} N$ is monotonic (for example, by differentiating), hence the unique answer is $N=5$.

share|cite|improve this answer

Note that $\;216 = 6^3,\;$ and $\;125 = 5^3,\;$ so use the logarithm power rule: $\;\log (a^b) = b \log a\;$ to write:

$$ (N+1)^{\large\log_N{ 5^3}} = (N+ 1)^{3\large\log_N5} = 6^3$$

That is $$\left [ (N+1)^3 \right ]^{\large\log_N{5}} = (5 + 1)^3 $$ and your solution is apparent.

share|cite|improve this answer

You can rewrite the equation so the solution is evident, such as

$$\left [ (N+1)^3 \right ]^{\log_N{5}} = (5 + 1)^3 $$

share|cite|improve this answer

I would first try to solve it by inspection, hoping that $N$ is something nice. I recognize $216$ as $6^3$, so I hope that $\log_N125=3$. I also recognize that $125=5^3$, so $\log_5 125=3$. And by great good fortune $6=5+1$, so $N=5$ is indeed a solution.

share|cite|improve this answer

Your equation is equal to finding an $N$ such that $(N+1)^{\log_N{5}}=6$. It is clear that if the natural number $N$ be greater than $5$ so $t=\log_N{5}<1$ and $(N+1)^{t}<6$. The same story is valid when $N<5$, because the function $x^t, t>1$ is increasing. So it would be $N=5$.

share|cite|improve this answer
@CalvinLin: Honestly, I considered $(N+1)^{\log_N^5}=6$ instead of $\left((N+1)^{\log_N^5}\right)^3=6^3$ which is the OP equation. Do you see any defects in it? Thanks. – Babak S. Jan 11 '13 at 17:18
@CalvinLin: I found out that the typo. Thanks Calvin. – Babak S. Jan 11 '13 at 18:28

Your Answer


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.