QUESTION 1
The given logarithm is
[tex]\log_{243}(27)[/tex]
Let [tex]\log_{243}(27)=x[/tex].
We rewrite in exponential form to get;
[tex]27=243^x[/tex]
We rewrite both sides of the equation as an index number to base 3.
[tex]3^3=3^{5x}[/tex]
Since the bases are the same, we equate the exponents.
[tex]3=5x[/tex]
Divide both sides by 5.
[tex]x=\frac{3}{5}[/tex]
[tex]\therefore \log_{243}(27)=\frac{3}{5}[/tex]
QUESTION 2
The given logarithm is
[tex]\log_{25}(\frac{1}{5} )[/tex]
We rewrite both the base and the number as power to base 5.
[tex]\log_{5^2}(5^{-1})[/tex]
Recall that: [tex]\log_{a^q}(a^p)=\frac{p}{q} \log_a(a)=\frac{p}{q}[/tex]
We apply this property to obtain;
[tex]\log_{5^2}(5^{-1})=\frac{-1}{2}\log_5(5)=-\frac{1}{2}[/tex]
