Q:

Evaluate each log without a calculator [tex]log_{243^{27} }[/tex][tex]log_{25} \frac{1}{5}[/tex]

Accepted Solution

A:
QUESTION 1The 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 2The 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]