I need to evaluate this log, i'm honestly not sure how to do one like this, even though i can do one without the fractions.

Answer:
- 5
Step-by-step explanation:
Using the law of logarithms
• [tex]log_{b}[/tex] x = n ⇔ x = [tex]b^{n}[/tex]
let [tex]log_{3}[/tex] [tex]\frac{1}{243}[/tex] = n, then
[tex]\frac{1}{243}[/tex] = [tex]3^{n}[/tex]
note that 243 = [tex]3^{5}[/tex], thus
[tex]3^{-5}[/tex] = [tex]3^{n}[/tex]
the bases on both sides are equal so equate the exponents
n = - 5
[tex]log_{3}[/tex] [tex]\frac{1}{243}[/tex] = - 5