Questions: Use properties of logarithms to expand the logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. log5(25/y) log5(25/y)=

Use properties of logarithms to expand the logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator.

log5(25/y)
log5(25/y)=

Transcript text: Use properties of logarithms to expand the logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. \[ \begin{array}{c} \log _{5}\left(\frac{25}{y}\right) \\ \log _{5}\left(\frac{25}{y}\right)= \end{array} \]

Solution

Solution Steps

Step 1: Apply the Quotient Rule of Logarithms

Given the expression: \[ \log_{5}\left(\frac{25}{y}\right) \] We apply the quotient rule of logarithms, which states: \[ \log_{b}\left(\frac{a}{c}\right) = \log_{b}(a) - \log_{b}(c) \] Thus, we get: \[ \log_{5}\left(\frac{25}{y}\right) = \log_{5}(25) - \log_{5}(y) \]

Step 2: Simplify \(\log_{5}(25)\)

Next, we simplify \(\log_{5}(25)\). Since \(25 = 5^2\), we can write: \[ \log_{5}(25) = \log_{5}(5^2) \] Using the power rule of logarithms, \(\log_{b}(a^c) = c \cdot \log_{b}(a)\), we get: \[ \log_{5}(5^2) = 2 \cdot \log_{5}(5) \] Since \(\log_{5}(5) = 1\), it simplifies to: \[ 2 \cdot 1 = 2 \]

Step 3: Substitute Back into the Expression

Substituting back, we have: \[ \log_{5}(25) - \log_{5}(y) = 2 - \log_{5}(y) \]