Prove $\log_5{30}<\log_8{81}$

It's easy to prove this by calculator or computer, and I wonder can we prove that $$\log_5{30}<\log_8{81}\tag 1$$ by pencil and paper ? Thanks in advance !

Edit: $(1)$ can be written as $$1+\log_52+\log_53<\frac{4}3\log_23,$$ but I can't go on with this.

In principle, sure. To show that $$\log_5 30<\frac{4216}{1995}<\log_8 81,$$ we only have to check $$30^{1995}<5^{4216}$$ and $$8^{4216}<81^{1995}.$$ Ok, not really helpful.

Ok,you can do it by changing bases of the logarithm. We have $\log_5{30}=\frac {\log_8{30}}{\log_8{5}}$ and we want to show that $$\log_5{30}=\frac {\log_8{30}}{\log_8{5}}<\log_8{81}\implies\\\log_8{30}<\log_8{81}\cdot \log_8{5}=\log_8{5}^{2\log_8{9}}\implies\\30<25^{\log_8{9}}$$

Now $\log_8{9}=\frac {\ln9}{\ln8}$ and show we need to show that $30^{\ln8}<25^{\ln9}$.

Let $f(x)=25^{\ln(x+1)}-30^{\ln x}$ then $f(x)>0$ for $x\in(0,8,0003)$ where at $8,0003$ we have the root. You can find the root by aplying Newton's method for finding roots.

For the proof we can use $\enspace\displaystyle \ln\frac{1+x}{1-x}= 2x \sum\limits_{k=0}^\infty\frac{x^{2k}}{2k+1}\enspace$ for $\,|x|<1\,$

and $\enspace \displaystyle \ln\frac{1+x}{1-x}< 2x \sum\limits_{k=0}^n\frac{x^{2k}}{2k+1}\enspace$ with $\enspace\displaystyle \sum\limits_{k=n+1}^\infty\frac{x^{2k}}{2k+1}<\frac{x^{2n+2}}{(2x+3)(1-x)}\,$ .

We have $\enspace\displaystyle\frac{1+x}{1-x}=2 => x=\frac{1}{3}\,$ , $\enspace\displaystyle\frac{1+x}{1-x}=3 => x=\frac{1}{2}\enspace$ and $\enspace\displaystyle\frac{1+x}{1-x}=5 => x=\frac{2}{3}\,$ .

Let’s choose an accuracy of $\,10^{-8}\,$ although it’s more than we need. This means:

$\ln 2\,:\enspace\displaystyle \frac{(1/3)^{2n+2}}{(2n+3)(1-1/3)}=\frac{1}{(4n+6)3^{2n+1}}<10^{-8}\enspace => \enspace \min(n)=7$

$\ln 3\,:\enspace\displaystyle \frac{(1/2)^{2n+2}}{(2n+3)(1-1/2)}=\frac{1}{(2n+3)2^{2n+1}}<10^{-8}\enspace => \enspace \min(n)=11$

$\ln 5\,:\enspace\displaystyle \frac{(2/3)^{2n+2}}{(2n+3)(1-2/3)}=\frac{1}{(n+1.5)1.5^{2n+1}}<10^{-8}\enspace => \enspace \min(n)=19$

It follows an equivalence for the question:

$\displaystyle 3(\ln 2)(\ln 2 + \ln 3 + \ln 5) < $ $\displaystyle < 3\left( \frac{2}{3} \left(\sum\limits_{k=0}^7\frac{(1/3)^{2k}}{2k+1} + \frac{(1/3)^{2\cdot 7 +2}}{(2\cdot 7 + 3)(1-1/3)}\right)\right)\left(\frac{2}{3}\left(\sum\limits_{k=0}^7\frac{(1/3)^{2k}}{2k+1} + \frac{(1/3)^{2\cdot 7 +2}}{(2\cdot 7 + 3)(1-1/3)}\right) + \frac{2}{2}\left(\sum\limits_{k=0}^{11}\frac{(1/2)^{2k}}{2k+1} + \frac{(1/2)^{2\cdot 11 +2}}{(2\cdot 11 + 3)(1-1/2)}\right) + \frac{4}{3} \left( \sum\limits_{k=0}^{19}\frac{(2/3)^{2k}}{2k+1} + \frac{(2/3)^{2\cdot 19 +2}}{(2\cdot 19 + 3)(1-2/3)}\right)\right) < 4 \left(\frac{2}{2} \sum\limits_{k=0}^{11}\frac{(1/2)^{2k}}{2k+1} \right)\left( \frac{4}{3} \sum\limits_{k=0}^{19}\frac{(2/3)^{2k}}{2k+1}\right) < 4(\ln 3)(\ln 5)$

We get (by hand, if you are patient)

$\displaystyle 3(\ln 2)(\ln 2 + \ln 3 + \ln 5)$

$\displaystyle <\enspace\frac{4961667635704152734727406651566353434163683633236701}{701534633600318577777221425020697457261844784742400}$

$\displaystyle <\enspace 7.072592 $

$\displaystyle <\enspace \frac{46645697273819775027503499400909017443549233759}{6595275179415105146772950235100943209168896000}$

$<\enspace 4(\ln 3)(\ln 5)$

which confirms the claim.