Question

b) $log_{3}\frac{1}{9}=$

Explanation:

Step1: Recall the logarithm definition

Let $y = \log_{3}\frac{1}{9}$. By the definition of logarithms, if $y=\log_{a}x$, then $a^{y}=x$. Here $a = 3$ and $x=\frac{1}{9}$, so $3^{y}=\frac{1}{9}$.

Step2: Rewrite $\frac{1}{9}$ as a power of 3

We know that $\frac{1}{9}=\frac{1}{3^{2}}=3^{- 2}$. So the equation $3^{y}=\frac{1}{9}$ becomes $3^{y}=3^{-2}$.

Step3: Equate the exponents

Since the bases are the same ($a = 3$), if $a^{m}=a^{n}$, then $m = n$. So for $3^{y}=3^{-2}$, we have $y=-2$.

Answer:

$-2$