Question

Expand log3 (h/9)

log3 (h-3)
Log3 h-2
Log3 ^h/2
(H-2)

Answer 1

To expand the expression \( \log_3 \left( \frac{h}{9} \right) \), we can use the properties of logarithms. The properties we will use are:

1. \( \log_b \left( \frac{x}{y} \right) = \log_b x - \log_b y \)
2. \( \log_b (x^n) = n \cdot \log_b x \)

Now, applying these properties:

\[
\log_3 \left( \frac{h}{9} \right) = \log_3 h - \log_3 9
\]

Next, we can simplify \( \log_3 9 \):

\[
\log_3 9 = \log_3 (3^2) = 2 \cdot \log_3 3 = 2
\]

Now we can substitute back:

\[
\log_3 \left( \frac{h}{9} \right) = \log_3 h - 2
\]

So, the expanded form of \( \log_3 \left( \frac{h}{9} \right) \) is:

\[
\log_3 h - 2
\]

### Summary
The expanded form of \( \log_3 \left( \frac{h}{9} \right) \) is:

\[
\log_3 h - 2
\]