MHB How can I simplify this logarithm?

[Elena1]

$$\log_{2}\left({24}\right) / \log_{96}\left({2}\right) - \log_{2}\left({192}\right) /\log_{12}\left({2}\right)$$

 

[I like Serena]

$$\log_{2}\left({24}\right) / \log_{96}\left({2}\right) - \log_{2}\left({192}\right) /\log_{12}\left({2}\right)$$

Hi Elena! :)

How far can you get?

Perhaps you can substitute the following? (Wondering)
$$\log_b y = \frac{\log_2 y}{\log_2 b}$$
That should make the expression simpler...

So for instance:
$$\log_{96} 2 = \frac{\log_2 2}{\log_2 96} = \frac{1}{\log_2 96}$$

 

[Elena1]

actually i can` t understand this subject can you explain more?please

 

[MarkFL]

I think what I like Serena meant is:

$$\log_{96}(2)=\frac{\log_2(2)}{\log_2(96)}=\frac{1}{\log_2(96)}$$

This is an application of the change of base formula.

 

[Elena1]

i know this...in the end of book i have the solution 3 but i obtained 0 the result

 

[I like Serena]

i know this...in the end of book i have the solution 3 but i obtained 0 the result

Good!
Can you show the first intermediate step? (Wondering)

 

[Elena1]

$$\log_{2}\left({6*{2}^{2}}\right) /\log_{6*{2}^{4}}\left({2}\right) -\log_{2}\left({3*{2}^{6}}\right)/ \log_{3*{2}^{2}}\left({2}\right)$$

- - - Updated - - -

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

 

[I like Serena]

$$\log_{2}\left({6*{2}^{2}}\right) /\log_{6*{2}^{4}}\left({2}\right) -\log_{2}\left({3*{2}^{6}}\right)/ \log_{3*{2}^{2}}\left({2}\right)$$

Good!

Can you change the base of each logarithm to $2$?

Btw, I have added $$$$ tags around your formula to properly render it.

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

That doesn't look quite right.
Let's zoom in on $\log_{6*{2}^{4}}\left({2}\right)$
It appears you deduced that it is equal to $\frac{1}{8}\log_{6}\left({2}\right)$

How did you get that? (Wondering)

 

[Elena1]

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

 

[MarkFL]

$$$$ \log_{6{2}^{4}}\left({2}\right)=\frac{1}{4}\log_{3*{2}^{2}}\left({2}\right)=\frac{1}{8}\log_{6}\left({2}\right)$$$$

You need to include $\LaTeX$ code within tags. On our toolbar you will see a button labeled with the $\Sigma$ character. Click this button and the MATH tags will be inserted at your current cursor location within the post message. Then, in between these tags (the cursor is already conveniently placed there) type your code. This will cause the code to be parsed correctly. :D

 

[I like Serena]

$$$$\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?

I see you have included $$$$ tags now.
However the $$ tag must go at the beginning, and the $$ must go at the end.

 

[I like Serena]

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

It should be:
$$\log_{6\cdot{2}^{4}}\left({2}\right)
=\frac{\log_6(2)}{\log_6(6\cdot {2}^{4})}
=\frac{\log_6(2)}{\log_6(6) + \log_6({2}^{4})}
=\frac{\log_6(2)}{1 + 4\log_6(2)}$$

Or better:
$$\log_{6\cdot{2}^{4}}\left({2}\right)
=\log_{3\cdot{2}^{5}}\left({2}\right)
=\frac{\log_2(2)}{\log_2(3\cdot {2}^{5})}
=\frac{1}{\log_2(3) + \log_2({2}^{5})}
=\frac{1}{\log_2(3) + 5}
$$