Prove that $\log_217\log_{\frac15}2\log_3\frac15>2$

Question

Prove that $\log_217\log_{\frac15}2\log_3\frac15>2$

I know that $\log_2 17>\log_216=4$. Don't know how to tinker with other factors.

Please put parentheses around all arguments, to make it clear what you're taking the logarithms of. For example, is it $\log_2(17 \log_{1/5} 2(\log_3(1/5)))?$ Or is it $\log_2(17) \log_{1/5}(2)\log_3(1/5)?$ — Adrian Keister, Aug 05 '19 at 18:20
Hi @AdrianKeister. I have posted the question as it was written in the book. I guess if it had meant the former, it would have used the parentheses. Without it, we would have to assume the latter, I guess. — aarbee, Aug 05 '19 at 18:22

score 5 · Answer 1 · answered Aug 05 '19 at 18:21

5

The left hand side $$=\dfrac{\log17\log2(-\log5)}{\log2(-\log5)(\log3)}=\log_317>\log_3(3^2)$$

answered Aug 05 '19 at 18:21

lab bhattacharjee

score 4 · Accepted Answer · answered Aug 05 '19 at 18:24

4

Write your product as $$\frac{\ln(17)}{\ln(2)}\cdot \frac{\ln(2)}{\ln\left(\frac{1}{5}\right)}\cdot\frac{ \ln\left(\frac{1}{5}\right)}{\ln(3)}$$

answered Aug 05 '19 at 18:24

score 1 · Answer 3 · answered Aug 05 '19 at 18:26

Note the following:

$\log_2 17 = \frac{\log 17}{\log 2}$ ; $\log_{\frac{1}{5}} 2 = -\frac{\log 2}{\log 5}$; $\log_3 \frac{1}{5} = -\frac{\log 5}{\log 3}$.

So multiplying together gives

$\log_2 17 \log_{\frac{1}{5}} 2 \log_3 \frac{1}{5} = \frac{\log 17}{\log 3} > \frac{\log 9}{\log 3} = 2$.

3 Answers3