Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 18306 by mondodotto@gmail.com last updated on 18/Jul/17

Answered by diofanto last updated on 18/Jul/17

$$\frac{\log 2^4-\log 3^4}{\log 3^3+\log 2^3}=\frac{4\log 2-4\log 3}{3\log 3+3\log 2}=$$$$=\frac{4(\log 2-\log 3)}{3(\log 2+\log 3)}=\frac{4\log \frac{2}{3}}{3\log 6}=\frac{4}{3}{\log }_6\left(\frac{2}{3}\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com