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Elena1
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\(\displaystyle (\sqrt{{\log_{n}\left({p}\right)} +\log_{p}\left({n}\right) +2} *(\log_{n}\left({p}\right)-\log_{np}\left({p}\right)) \sqrt{\log_{n}\left({p}\right)}\)
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Elena said:\(\displaystyle (\sqrt{\log_{n}\left({p}\right)} +\log_{p}\left({n}\right) +2) *(\log_{n}\left({p}\right)-\log_{np}\left({p}\right)) \sqrt{\log_{n}\left({p}\right)}\)
Elena said:under the radical is the second expresion, also
Elena said:i did it
Elena said:\(\displaystyle (\sqrt{{\log_{n}\left({p}\right)} +\log_{p}\left({n}\right) +2} *(\log_{n}\left({p}\right)-\log_{np}\left({p}\right)) \sqrt{\log_{n}\left({p}\right)}\)
Elena said:\(\displaystyle \sqrt{\frac{1}{\log_{p}\left({n}\right)} +\log_{p}\left({n}\right)}+2} * \frac{1}{\log_{p}\left({n}\right)-\frac{1}{\log_{p}\left({np}\right)}* \sqrt{\frac{1}{\log_{p}\left({n}\right)}}\)
Elena said:what ?
Elena said:how to bring the same denominator? what should I receive at the top(numerator)
I like Serena said:Suppose you want to calculate $\frac 1 2 - \frac 1 5$.
What does the top become? (Wondering)
Elena said:3/10
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can you help me please to solve thoroughly?
I like Serena said:Yes.
How did you find 3/10 exactly?
That's what we want to do with \(\displaystyle \frac{1}{\log_{p}({n})}-\frac{1}{\log_{p}({n}) + 1}\) as well...
Elena said:i know but i don`t understand it with logariths can you show/
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)
Elena said:\(\displaystyle \frac{\log_{p}\left({n+1}\right)-\log_{p}\left({n}\right)}{\log_{p}\left({n}\right)*\log_{p}\left({n+1}\right)}\)
Elena said:\(\displaystyle \frac{\log_{p}\left({n+1}\right)-\log_{p}\left({n}\right)}{\log_{p}\left({n}\right)*\log_{p}\left({n+1}\right)}\)
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\(\displaystyle \frac{1}{\log_{p}\left({n}\right)*\log_{p}\left({n+1}\right)}\)
and \(\displaystyle \log_{p}\left({n}\right)*\log_{p}\left({n}\right)= 2\log_{p}\left({n}\right)\)I like Serena said:Close!
But your parentheses are not quite right.
It should be:
$$\frac 1{\log_p(n)} - \frac 1{\log_p(n) + 1}
=\frac{(\log_p(n) + 1)-\log_{p}\left({n}\right)}{\log_{p}\left({n}\right)\cdot(\log_p(n) + 1)}
=\frac{1}{\log_{p}\left({n}\right)\cdot(\log_p(n) + 1)}
$$
Elena said:and \(\displaystyle \log_{p}\left({n}\right)*\log_{p}\left({n}\right)= 2\log_{p}\left({n}\right)\)
A logarithm is a mathematical function that is the inverse of the exponential function. It is used to solve exponential equations and represents the power or exponent to which a base number must be raised to obtain a given value.
Simplifying a logarithm means to rewrite it in its most compact form, by using logarithm rules and properties to combine terms and reduce complex expressions into simpler ones. This makes it easier to solve equations and perform calculations.
The basic rules for simplifying logarithms include the product rule, quotient rule, power rule, and change of base rule. These rules can be used to combine terms, move exponents, and change the base of a logarithm to a more convenient one.
To simplify a logarithm with multiple terms, you can use the product rule or quotient rule to combine the terms into a single logarithm. Then, you can use the power rule to move the exponent of the resulting logarithm to the front, simplifying the expression further.
Not all logarithms can be simplified, as some may already be in their most simplified form. Additionally, some logarithms may be unable to be simplified due to the values of the base and argument, resulting in an irrational or undefined answer.