MHB What is the relationship between logarithms and simplifying expressions?

AI Thread Summary
The discussion focuses on the relationship between logarithms and simplifying expressions, particularly how to change logarithmic bases and simplify complex expressions. Participants explore converting logarithms to a common base, specifically base \( p \), and discuss the implications of this transformation on the expressions provided. They emphasize the importance of making denominators equal when subtracting fractions involving logarithms, drawing parallels to basic arithmetic operations. The conversation includes specific examples and calculations to illustrate the simplification process, ultimately aiming to clarify the mathematical principles involved. The thread highlights the significance of understanding logarithmic properties for effective expression simplification.
Elena1
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$$(\sqrt{{\log_{n}\left({p}\right)} +\log_{p}\left({n}\right) +2} *(\log_{n}\left({p}\right)-\log_{np}\left({p}\right)) \sqrt{\log_{n}\left({p}\right)}$$
 
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Elena said:
$$(\sqrt{\log_{n}\left({p}\right)} +\log_{p}\left({n}\right) +2) *(\log_{n}\left({p}\right)-\log_{np}\left({p}\right)) \sqrt{\log_{n}\left({p}\right)}$$

What do you get if you change everything to the same base? (Wondering)
 
under the radical is the second expresion, also
 
Elena said:
under the radical is the second expresion, also

Please post it as it should be.
 
i did it
 
Elena said:
i did it

Good! ;)

Can you change the logarithms to the same base now?
Let's say to base $p$?

For instance:
$$\log_n(p) = \frac{\log_p(p)}{\log_p(n)}$$
 
$$\sqrt{\frac{1}{\log_{p}\left({n}\right)} +\log_{p}\left({n}\right)}+2} * \frac{1}{\log_{p}\left({n}\right)-\frac{1}{\log_{p}\left({np}\right)}* \sqrt{\frac{1}{\log_{p}\left({n}\right)}}$$
 
Elena said:
$$(\sqrt{{\log_{n}\left({p}\right)} +\log_{p}\left({n}\right) +2} *(\log_{n}\left({p}\right)-\log_{np}\left({p}\right)) \sqrt{\log_{n}\left({p}\right)}$$

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

I guess you mean:
$$\sqrt{\frac{1}{\log_{p}\left({n}\right)} +\log_{p}\left({n}\right) + 2}
\cdot \left(\frac{1}{\log_{p}\left({n}\right)}-\frac{1}{\log_{p}\left({np}\right)}\right)
\cdot \sqrt{\frac{1}{\log_{p}\left({n}\right)}}$$
:rolleyes:

How about simplifying a little? (Wondering)
 
what ?
 
  • #10
Elena said:
what ?

Well... $\log_p(np)=\log_p(n) + \log_p(p)=\log_p(n) + 1$.
 
  • #11
how to bring the same denominator? what should I receive at the top(numerator)
 
  • #12
Elena said:
how to bring the same denominator? what should I receive at the top(numerator)

Suppose you want to calculate $\frac 1 2 - \frac 1 5$.
What does the top become? (Wondering)
 
  • #13
I like Serena said:
Suppose you want to calculate $\frac 1 2 - \frac 1 5$.
What does the top become? (Wondering)

3/10

- - - Updated - - -

can you help me please to solve thoroughly?
 
  • #14
Elena said:
3/10

- - - Updated - - -

can you help me please to solve thoroughly?

Yes.

How did you find 3/10 exactly?

That's what we want to do with $$\frac{1}{\log_{p}({n})}-\frac{1}{\log_{p}({n}) + 1}$$ as well...
 
  • #15
I like Serena said:
Yes.

How did you find 3/10 exactly?

That's what we want to do with $$\frac{1}{\log_{p}({n})}-\frac{1}{\log_{p}({n}) + 1}$$ as well...

i know but i don`t understand it with logariths can you show/
 
  • #16
Elena said:
i know but i don`t understand it with logariths can you show/

Making it explicit, we have:
$$\frac 1 2 - \frac 1 5
= \frac {1\cdot 5} {2\cdot 5} - \frac {1\cdot 2} {2\cdot 5}
= \frac {1\cdot 5 - 1\cdot 2} {2\cdot 5}
=\frac{3}{10}
$$
First we make the denominators equal, then we subtract.

More generally:
$$\frac a b - \frac c d
= \frac {a\cdot d - b\cdot c} {b\cdot d}
$$

Can you apply it to the fractions with the logarithms? (Wondering)
 
  • #17
I like Serena said:
Making it explicit, we have:
$$\frac 1 2 - \frac 1 5
= \frac {1\cdot 5} {2\cdot 5} - \frac {1\cdot 2} {2\cdot 5}
= \frac {1\cdot 5 - 1\cdot 2} {2\cdot 5}
=\frac{3}{10}
$$
First we make the denominators equal, then we subtract.

More generally:
$$\frac a b - \frac c d
= \frac {a\cdot d - b\cdot c} {b\cdot d}
$$

Can you apply it to the fractions with the logarithms? (Wondering)

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

- - - Updated - - -

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

$$\frac{1}{\log_{p}\left({n}\right)*\log_{p}\left({n+1}\right)}$$
 
  • #18
Elena said:
$$\frac{\log_{p}\left({n+1}\right)-\log_{p}\left({n}\right)}{\log_{p}\left({n}\right)*\log_{p}\left({n+1}\right)}$$

- - - Updated - - -

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

Close!

But your parentheses are not quite right.
It should be:
$$\frac 1{\log_p(n)} - \frac 1{\log_p(n) + 1}
=\frac{(\log_p(n) + 1)-\log_{p}\left({n}\right)}{\log_{p}\left({n}\right)\cdot(\log_p(n) + 1)}
=\frac{1}{\log_{p}\left({n}\right)\cdot(\log_p(n) + 1)}
$$
 
  • #19
I like Serena said:
Close!

But your parentheses are not quite right.
It should be:
$$\frac 1{\log_p(n)} - \frac 1{\log_p(n) + 1}
=\frac{(\log_p(n) + 1)-\log_{p}\left({n}\right)}{\log_{p}\left({n}\right)\cdot(\log_p(n) + 1)}
=\frac{1}{\log_{p}\left({n}\right)\cdot(\log_p(n) + 1)}
$$
and $$\log_{p}\left({n}\right)*\log_{p}\left({n}\right)= 2\log_{p}\left({n}\right)$$
 
  • #20
Elena said:
and $$\log_{p}\left({n}\right)*\log_{p}\left({n}\right)= 2\log_{p}\left({n}\right)$$

Sorry, but no.
It should be:
$$\log_{p}\left({n}\right)\cdot\log_{p}\left({n}\right)= (\log_{p}(n))^2$$

I suspect you were thinking of:
$$\log_{p}(n\cdot n) = 2\log_{p}({n})$$
But that is something different.
 

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