What is the relationship between logarithms and simplifying expressions?

[Elena1]

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

 

[I like Serena]

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

What do you get if you change everything to the same base? (Wondering)

 

[Elena1]

under the radical is the second expresion, also

 

[MarkFL]

under the radical is the second expresion, also

Please post it as it should be.

 

[Elena1]

i did it

 

[I like Serena]

i did it

Good! ;)

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

For instance:
$$\log_n(p) = \frac{\log_p(p)}{\log_p(n)}$$

 

[Elena1]

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

 

[I like Serena]

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

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

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


How about simplifying a little? (Wondering)

 

[Elena1]

what ?

 

[I like Serena]

what ?

Well... $\log_p(np)=\log_p(n) + \log_p(p)=\log_p(n) + 1$.

 

[Elena1]

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

 

[I like Serena]

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)

 

[Elena1]

Suppose you want to calculate $\frac 1 2 - \frac 1 5$.
What does the top become? (Wondering)

3/10

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can you help me please to solve thoroughly?

 

[I like Serena]

3/10

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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...

 

[Elena1]

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/

 

[I like Serena]

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)

 

[Elena1]

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)

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

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

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

 

[I like Serena]

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

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

Close!

But your parentheses are not quite right.
It should be:
$$\frac 1{\log_p(n)} - \frac 1{\log_p(n) + 1}
=\frac{(\log_p(n) + 1)-\log_{p}\left({n}\right)}{\log_{p}\left({n}\right)\cdot(\log_p(n) + 1)}
=\frac{1}{\log_{p}\left({n}\right)\cdot(\log_p(n) + 1)}
$$

 

[Elena1]

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

 

[I like Serena]

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.