Expand the following logarithmic expression:

[tex]\[ \log_5\left(8 \frac{\sqrt{t}}{v}\right) \][/tex]

A. [tex]\(\log_5 8 + 2 \log_5 t - \log_5 v\)[/tex]

B. [tex]\(\log_5 8 + \log_5 \frac{1}{2} t - \log_5 v\)[/tex]

C. [tex]\(\frac{1}{2} \log_5 8 + \frac{1}{2} \log_5 t - \frac{1}{2} \log_5 v\)[/tex]

D. [tex]\(\log_5 8 + \frac{1}{2} \log_5 t - \log_5 v\)[/tex]



Answer :

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.

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