Explanations:
Given the following parameters:
log 5 = A
log 3 = B
According to the law of product and quotient of logarithm as shown:
[tex]\begin{gathered} \log AB=\log A+\log B \\ \log (\frac{A}{B})=\log A-\log B \\ \log A^b=b\log A \end{gathered}[/tex]
Applying the laws of logarithm in solving the given logarithm
[tex]\begin{gathered} a)\log 15 \\ =\log (5\times3) \\ =\log 5+\log 3 \\ =A+B \end{gathered}[/tex]
For the expression log(25/3)
[tex]\begin{gathered} b)\log (\frac{25}{3}) \\ =\log (\frac{5^2}{3}) \\ =\log 5^2-\log 3 \\ =2\log 5-\log 3 \\ =2A-B \end{gathered}[/tex]
For the expression log135
[tex]\begin{gathered} \log (135) \\ =\log (5\times27) \\ =\log (5^{}\times3^3) \\ =\log 5^{}+\log 3^3 \\ =\log 5+3\log 3 \\ =A+3B \end{gathered}[/tex]
For the expression log₅27
[tex]\begin{gathered} \log _527 \\ =\frac{\log 27}{\log 5} \\ =\frac{\log 3^3}{\log 5} \\ =\frac{3\log 3}{\log 5} \\ =\frac{3B}{A} \end{gathered}[/tex]
For the expression log₉625
[tex]\begin{gathered} \log _9625 \\ =\frac{\log 625}{\log 9} \\ =\frac{\log 5^4}{\log 3^2} \\ =\frac{4\log 5}{2\log 3} \\ =\frac{\cancel{4}^2A}{\cancel{2}B} \\ =\frac{2A}{B} \end{gathered}[/tex]
For the value of 15, this can be expressed as shown. Since:
[tex]\begin{gathered} \log 5=A;10^A=5 \\ \log 3=B;10^B=3^{} \end{gathered}[/tex]
Since 15 = 5 × 3, writing it in terms of A and B will be expressed as:
[tex]\begin{gathered} 15=5\times3 \\ 15=10^A\times10^B \\ 15=10^{A+B} \end{gathered}[/tex]
