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# How do you write the logarithmic expression as the sum, difference or multiple of logarithms and simplify as much as possible for ${\log _5}\left( {\dfrac{{25}}{x}} \right)$?

Last updated date: 09th Aug 2024
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Hint: The above question is based on the concept of logarithm. The main approach towards solving the question is by applying various logarithmic properties on the above given expression. Properties like quotient rule and power rule are applied to the expression and further it is simplified.

Complete step by step solution:
Logarithms, like exponents, have many useful properties that can be used to reduce by simplifying logarithmic expressions and solve logarithmic equations. Logarithm is the exponent or power to which base must be raised to yield a given number. When expressed mathematically x is the logarithm of base n to the base b if ${b^x} = n$ ,then we can write it as $x = {\log _b}n$.

The above given expression is ${\log _5}\left( {\dfrac{{25}}{x}} \right)$.So first we will apply quotient rule which is given below:
${\log _b}\left( {\dfrac{M}{N}} \right) = {\log _b}M - {\log _b}N$
Now this states that the log of quotients is the difference of the log of dividend and divisor. So, applying this on the expression,
$\Rightarrow \log \left( {\dfrac{{25}}{x}} \right) = {\log _5}\left( {25} \right) - {\log _5}x \\ \Rightarrow {\log _5}\left( {\dfrac{{25}}{x}} \right) = {\log _5}\left( {{5^2}} \right) - {\log _5}x \\$
Now by applying the power rule which is given below:
${\log _b}\left( {{M^p}} \right) = p{\log _b}\left( M \right)$
Since we have the first term in power therefore will apply the other property.
${\log _5}\left( {\dfrac{{25}}{x}} \right) = {\log _5}\left( {{5^2}} \right) - \log x \\ \Rightarrow{\log _5}\left( {\dfrac{{25}}{x}} \right) = 2{\log _5}\left( 5 \right) - \log x \\ \therefore{\log _5}\left( {\dfrac{{25}}{x}} \right) = (2 \times 1) - {\log _5}x = 2 - {\log _5}x$
Therefore, we get the above solution after simplifying.

Note: An important thing to note is that these properties can have any values for M,N and b where $M,N > 0$and $0 < b \ne 1$.The reason for this is the argument of the logarithm must be positive and the base of the logarithm must also be positive and not equal to 1.