State whether $\log_23 < \log_517$. 
State whether true or false :
$\log_23 < \log_517$

I have tried to make a common base and then proceed but I am stuck , I am not able to figure out how can I predict this without using a calculator.
Could someone please guide me with some hint on how to compare logarithmic quantities with different bases ?
Thanks
 A: I claim $\log_2(3) < 5/3 < \log_5(17)$.
The first is equivalent to $2^{5/3} > 3$, i.e. $2^5 > 3^3$.  That's easy: $2^5 = 32$ and $3^3 = 27$.
The second is equivalent to $5^{5/3} < 17$, i.e.
$5^5 = 3125 < 4913 = 17^3$.  That's rather tedious, but doable, if computing by hand.
A: As there are now several explicit answers (there were none before), I'll expand my earlier comment:
$$
\log_5{17} > \log_5{15} = 1 + \log_53 > 1 + \frac23 = \frac53 > \log_23,
$$
because
$$
3^3 = 27 > 25 = 5^2
$$
and
$$
2^5 = 32 > 27 = 3^3.
$$
A: $$\log^a_b = \frac{\ln^a}{\ln^b}$$
In the following, if $a,b,c,d$ are all positive, then 
$$\frac{a}{b} < \frac{c}{d} \rightarrow ad< bc$$
https://brilliant.org/wiki/does-cross-multiple-always-work-for-inequalities/
Also,
$$a<b \iff \ln^{a}< \ln^{b}$$

So your inequality becomes,
$$\frac{\ln^3}{\ln^2} < \frac{\ln^{17}}{\ln^{5}}$$
And now you have everything in the same base.
I would leave it at this for you to solve the rest. If you still have problems, then let me know in the comments. 
A: To prove that:
$$\log_2 3 < \log_5 17 \implies \log_2 3< \frac{\log_2 17}{\log_2 5} ~~~(1)$$
$$17 > 2^4 \implies \log_2 17 >4, ~~~(2)$$ $$ 25 < 32 \implies 5 <2^{5/2} \implies \log_2 5 < \frac{5}{2}~~~(3) $$ 
Writing (2) and (3) inte same direction and multiplying them, we get
$$\log_5 17 >\frac{8}{5}~~~~~~(4)$$
Next from $$ 243 <256 \implies 3^5 < 2^8 \implies 3< 2^{8/5} \implies \log_2 3 <\frac{8}{5}~~~(5).$$
Finlly (4) and (5) prove (1).
