How can I simplify this logarithm?

In summary, you can deduce that $\log_{6*{2}^{4}}\left({2}\right)$ is equal to $\frac{1}{4}\log_{3*{2}^{2}}\left({2}\right)$
  • #1
Elena1
24
0
\(\displaystyle \log_{2}\left({24}\right) / \log_{96}\left({2}\right) - \log_{2}\left({192}\right) /\log_{12}\left({2}\right)\)
 
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  • #2
Elena said:
\(\displaystyle \log_{2}\left({24}\right) / \log_{96}\left({2}\right) - \log_{2}\left({192}\right) /\log_{12}\left({2}\right)\)

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:
$$\log_{96} 2 = \frac{\log_2 2}{\log_2 96} = \frac{1}{\log_2 96}$$
 
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  • #3
actually i can` t understand this subject can you explain more?please
 
  • #4
I think what I like Serena meant is:

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

This is an application of the change of base formula.
 
  • #5
i know this...in the end of book i have the solution 3 but i obtained 0 the result
 
  • #6
Elena said:
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)
 
  • #7
\(\displaystyle \log_{2}\left({6*{2}^{2}}\right) /\log_{6*{2}^{4}}\left({2}\right) -\log_{2}\left({3*{2}^{6}}\right)/ \log_{3*{2}^{2}}\left({2}\right)\)

- - - Updated - - -

finally i obtained \(\displaystyle 4 \log_{2}\left({6}\right) / \frac{1}{8}\log_{6}\left({2}\right) -6\log_{2}\left({6}\right) /
\frac{1}{2}\log_{6}\left({2}\right)\)
 
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  • #8
Elena said:
\(\displaystyle \log_{2}\left({6*{2}^{2}}\right) /\log_{6*{2}^{4}}\left({2}\right) -\log_{2}\left({3*{2}^{6}}\right)/ \log_{3*{2}^{2}}\left({2}\right)\)

Good!

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

Btw, I have added [MATH][/MATH] tags around your formula to properly render it.

Elena said:
finally i obtained \(\displaystyle 4 \log_{2}\left({6}\right) / \frac{1}{8}\log_{6}\left({2}\right) -6\log_{2}\left({6}\right) /
\frac{1}{2}\log_{6}\left({2}\right)\)

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)
 
  • #9
\(\displaystyle \log_{6*{2}^{4}}\left({2}\right)=\frac{1}{4}\log_{3*{2}^{2}}\left({2}\right)=\frac{1}{8}\log_{6}\left({2}\right)\)
how can i write to be clear?
 
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  • #10
Elena said:
\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
 
  • #11
Elena said:
\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)
Code:
how can i write to be clear?

I see you have included [MATH][/MATH] tags now.
However the [MATH] tag must go at the beginning, and the [/MATH] must go at the end.
 
  • #12
Elena said:
\(\displaystyle \log_{6*{2}^{4}}\left({2}\right)=\frac{1}{4}\log_{3*{2}^{2}}\left({2}\right)=\frac{1}{8}\log_{6}\left({2}\right)\)
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:
$$\log_{6\cdot{2}^{4}}\left({2}\right)
=\log_{3\cdot{2}^{5}}\left({2}\right)
=\frac{\log_2(2)}{\log_2(3\cdot {2}^{5})}
=\frac{1}{\log_2(3) + \log_2({2}^{5})}
=\frac{1}{\log_2(3) + 5}
$$
 

FAQ: How can I simplify this logarithm?

What is a logarithm?

A logarithm is the inverse function of an exponential function. It is used to solve exponential equations and is written as logb(x) where b is the base and x is the argument or input of the logarithm.

Why do we need to simplify logarithms?

Simplifying logarithms makes them easier to work with and helps to solve complex equations. It also helps to identify patterns and relationships between numbers.

What are the rules for simplifying logarithms?

The main rules for simplifying logarithms are the product rule, quotient rule, and power rule. The product rule states that logb(xy) = logb(x) + logb(y). The quotient rule states that logb(x/y) = logb(x) - logb(y). And the power rule states that logb(xn) = nlogb(x).

What are some common mistakes when simplifying logarithms?

Some common mistakes when simplifying logarithms include forgetting to apply the rules correctly, using the wrong base, and not simplifying the final answer. It is important to double-check your work and simplify as much as possible.

How do we simplify logarithms with different bases?

To simplify logarithms with different bases, we can use the change of base formula: logb(x) = logc(x)/logc(b), where c is any base. This allows us to convert the logarithm to a common base and then apply the rules for simplifying.

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