Recent content by Elena1

could you solve my problem please i don`t understand the drawing

я не знаю

I mean the distance between the intersection point of bisectors and the intersection point of medians

The legs of chateti of a right triangle are 9 and 12 cm. Find the distance between the intersection point of bisectors and the point of intersection of the medians

and $$\log_{p}\left({n}\right)*\log_{p}\left({n}\right)= 2\log_{p}\left({n}\right)$$

$$\frac{\log_{p}\left({n+1}\right)-\log_{p}\left({n}\right)}{\log_{p}\left({n}\right)*\log_{p}\left({n+1}\right)}$$ - - - Updated - - - $$\frac{1}{\log_{p}\left({n}\right)*\log_{p}\left({n+1}\right)}$$

i know but i don`t understand it with logariths can you show/

3/10 - - - Updated - - - can you help me please to solve thoroughly?

how to bring the same denominator? what should I receive at the top(numerator)

what ?

$$\sqrt{\frac{1}{\log_{p}\left({n}\right)} +\log_{p}\left({n}\right)}+2} * \frac{1}{\log_{p}\left({n}\right)-\frac{1}{\log_{p}\left({np}\right)}* \sqrt{\frac{1}{\log_{p}\left({n}\right)}}$$

i did it

$$\log_{6*{2}^{4}}\left({2}\right)=\frac{1}{4}\log_{3*{2}^{2}}\left({2}\right)=\frac{1}{8}\log_{6}\left({2}\right)$$ how can i write to be clear?

under the radical is the second expresion, also

$$\log_{2}\left({6*{2}^{2}}\right) /\log_{6*{2}^{4}}\left({2}\right) -\log_{2}\left({3*{2}^{6}}\right)/ \log_{3*{2}^{2}}\left({2}\right)$$ - - - Updated - - - finally i obtained $$4 \log_{2}\left({6}\right) / \frac{1}{8}\log_{6}\left({2}\right) -6\log_{2}\left({6}\right) /...