The Minimum Possible Base For Two Logs

$\log_{5n} 30\sqrt{5} \ge \log_{4n} 48 \Rightarrow 30\sqrt{5} \ge (5n)^{\log_{4n} 48}$ .

Now $30\sqrt{5} \ge (\frac{5}{4} \times 4n)^{\log_{4n} 48} =(\frac{5}{4})^{\log_{4n} 48}(4n)^{\log_{4n} 48}= (\frac{5}{4})^{\log_{4n} 48} \times 48$ .

This implies that $\frac{5\sqrt{5}}{8} \ge (\frac{5}{4})^{\log_{4n} 48}$ . Note that $\frac{5\sqrt{5}}{8} = (\frac{5}{4})^{\frac{3}{2}}$ , hence $(\frac{5}{4})^{\frac{3}{2}} \ge (\frac{5}{4})^{\log_{4n} 48}$ which means that $\frac{3}{2}\ge \log_{4n} 48$ .

Thus $(4n)^{\frac{3}{2}}\ge 48 \Rightarrow 8 n^{\frac{3}{2}}\ge 48 \Rightarrow n^{\frac{3}{2}}\ge 6 \Rightarrow n^3\ge 36$ .

So the minimum value of $n^3$ is $\color{#3D99F6} {36}$ .

I used change of base to find the maximum lower bound, hence $M$

Let's change the base to 10 ( so log from now on means $log_{10}$ ).

We get at equality that

$\frac{log (30 \sqrt{5})}{log(5n)} = \frac{ log(48)}{log(4n)}$

or

$log(30 \sqrt{5}) . log(4n) = log (48) . log(5n)$ .

Using the identity that $log(a.b) = log(a) + log(b)$ , we get

$log(30 \sqrt{5}) . (log (4) + log(n)) = log (48 ). (log (5) + log(n))$ .

Isolating log n, we obtain :

$log(n) = \dfrac{ log(30 \sqrt{5}) . log (4) - log (48) . log (5)}{log (48) - log (30 \sqrt{5}) }$ ,

which gives the solution $n^3 = 36$ , using a pocket calculator.