State whether $\log_23 < \log_517$.

Question

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

Answer 1

$$\log^a_b = \frac{\ln^a}{\ln^b}$$ 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,

$$a<b \iff \ln^{a}< \ln^{b}$$

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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.

Answer 2

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.

Answer 3

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).

Answer 4

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 $$ 2^5 = 32 > 27 = 3^3. $$