Certainly! Let's go step by step to expand the given logarithmic expression:
[tex]$
\log_5\left(8 \frac{\sqrt{t}}{v}\right)
$[/tex]
### Step 1: Separate the Product Rule
We use the property of logarithms that states:
[tex]$
\log_b(x \cdot y) = \log_b(x) + \log_b(y)
$[/tex]
Here, [tex]\( x = 8 \)[/tex] and [tex]\( y = \frac{\sqrt{t}}{v} \)[/tex], so:
[tex]$
\log_5\left(8 \cdot \frac{\sqrt{t}}{v}\right) = \log_5(8) + \log_5\left(\frac{\sqrt{t}}{v}\right)
$[/tex]
### Step 2: Separate the Quotient Rule
Next, we use the property of logarithms that states:
[tex]$
\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)
$[/tex]
Here, [tex]\( x = \sqrt{t} \)[/tex] and [tex]\( y = v \)[/tex], so:
[tex]$
\log_5\left(\frac{\sqrt{t}}{v}\right) = \log_5(\sqrt{t}) - \log_5(v)
$[/tex]
Combining this with the first step, we get:
[tex]$
\log_5(8) + \log_5(\sqrt{t}) - \log_5(v)
$[/tex]
### Step 3: Use the Power Rule
Finally, we use the property of logarithms that states:
[tex]$
\log_b(x^r) = r \log_b(x)
$[/tex]
Here, [tex]\( \sqrt{t} = t^{1/2} \)[/tex], so:
[tex]$
\log_5(\sqrt{t}) = \log_5(t^{1/2}) = \frac{1}{2} \log_5(t)
$[/tex]
Putting it all together, we get:
[tex]$
\log_5(8) + \frac{1}{2} \log_5(t) - \log_5(v)
$[/tex]
So, the final expanded expression for
[tex]$
\log_5\left(8 \frac{\sqrt{t}}{v}\right)
$[/tex]
is:
[tex]$
\log_5(8) + \frac{1}{2} \log_5(t) - \log_5(v)
$[/tex]
This matches the final form you are given.
