Solving an equation with a logarithm in the exponent

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?

Thanks!

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 $N=5$.

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.

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

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

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.

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 $N=5$.

• @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. – mrs Jan 11 '13 at 17:18
• @CalvinLin: I found out that the typo. Thanks Calvin. – mrs Jan 11 '13 at 18:28