Answer :
To expand the logarithmic expression [tex]\(\log_5\left(8 \frac{\sqrt{t}}{v}\right)\)[/tex], we will use the properties of logarithms. Specifically, we will apply the following properties of logarithms:
1. [tex]\(\log_b(xy) = \log_b(x) + \log_b(y)\)[/tex]
2. [tex]\(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\)[/tex]
3. [tex]\(\log_b(x^c) = c \log_b(x)\)[/tex]
Given the expression [tex]\(\log_5\left(8 \frac{\sqrt{t}}{v}\right)\)[/tex]:
1. We start by recognizing that the argument inside the logarithm, [tex]\(8 \frac{\sqrt{t}}{v}\)[/tex], is a product of two parts: [tex]\(8\)[/tex] and [tex]\(\frac{\sqrt{t}}{v}\)[/tex].
2. Using the product property ([tex]\(\log_b(xy) = \log_b(x) + \log_b(y)\)[/tex]), we can write:
[tex]\[ \log_5\left(8 \frac{\sqrt{t}}{v}\right) = \log_5(8) + \log_5\left(\frac{\sqrt{t}}{v}\right) \][/tex]
3. Next, we deal with the term [tex]\(\frac{\sqrt{t}}{v}\)[/tex] using the quotient property ([tex]\(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\)[/tex]):
[tex]\[ \log_5\left(\frac{\sqrt{t}}{v}\right) = \log_5(\sqrt{t}) - \log_5(v) \][/tex]
4. Now, we need to expand the [tex]\(\log_5(\sqrt{t})\)[/tex] term. Realize that [tex]\(\sqrt{t}\)[/tex] can be rewritten as [tex]\(t^{1/2}\)[/tex]. Using the power property ([tex]\(\log_b(x^c) = c \log_b(x)\)[/tex]), we get:
[tex]\[ \log_5(\sqrt{t}) = \log_5(t^{1/2}) = \frac{1}{2} \log_5(t) \][/tex]
5. Putting it all together, we substitute back into our original expression:
[tex]\[ \log_5\left(8 \frac{\sqrt{t}}{v}\right) = \log_5(8) + \left(\frac{1}{2} \log_5(t) - \log_5(v)\right) \][/tex]
6. Finally, we combine all the terms:
[tex]\[ \log_5\left(8 \frac{\sqrt{t}}{v}\right) = \log_5(8) + \frac{1}{2} \log_5(t) - \log_5(v) \][/tex]
Therefore, the expanded expression is:
[tex]\[ \log_5 8 + \frac{1}{2} \log_5 t - \log_5 v \][/tex]
So, the correct expansion is [tex]\(\log_5 8 + \frac{1}{2} \log_5 t - \log_5 v\)[/tex].
1. [tex]\(\log_b(xy) = \log_b(x) + \log_b(y)\)[/tex]
2. [tex]\(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\)[/tex]
3. [tex]\(\log_b(x^c) = c \log_b(x)\)[/tex]
Given the expression [tex]\(\log_5\left(8 \frac{\sqrt{t}}{v}\right)\)[/tex]:
1. We start by recognizing that the argument inside the logarithm, [tex]\(8 \frac{\sqrt{t}}{v}\)[/tex], is a product of two parts: [tex]\(8\)[/tex] and [tex]\(\frac{\sqrt{t}}{v}\)[/tex].
2. Using the product property ([tex]\(\log_b(xy) = \log_b(x) + \log_b(y)\)[/tex]), we can write:
[tex]\[ \log_5\left(8 \frac{\sqrt{t}}{v}\right) = \log_5(8) + \log_5\left(\frac{\sqrt{t}}{v}\right) \][/tex]
3. Next, we deal with the term [tex]\(\frac{\sqrt{t}}{v}\)[/tex] using the quotient property ([tex]\(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\)[/tex]):
[tex]\[ \log_5\left(\frac{\sqrt{t}}{v}\right) = \log_5(\sqrt{t}) - \log_5(v) \][/tex]
4. Now, we need to expand the [tex]\(\log_5(\sqrt{t})\)[/tex] term. Realize that [tex]\(\sqrt{t}\)[/tex] can be rewritten as [tex]\(t^{1/2}\)[/tex]. Using the power property ([tex]\(\log_b(x^c) = c \log_b(x)\)[/tex]), we get:
[tex]\[ \log_5(\sqrt{t}) = \log_5(t^{1/2}) = \frac{1}{2} \log_5(t) \][/tex]
5. Putting it all together, we substitute back into our original expression:
[tex]\[ \log_5\left(8 \frac{\sqrt{t}}{v}\right) = \log_5(8) + \left(\frac{1}{2} \log_5(t) - \log_5(v)\right) \][/tex]
6. Finally, we combine all the terms:
[tex]\[ \log_5\left(8 \frac{\sqrt{t}}{v}\right) = \log_5(8) + \frac{1}{2} \log_5(t) - \log_5(v) \][/tex]
Therefore, the expanded expression is:
[tex]\[ \log_5 8 + \frac{1}{2} \log_5 t - \log_5 v \][/tex]
So, the correct expansion is [tex]\(\log_5 8 + \frac{1}{2} \log_5 t - \log_5 v\)[/tex].
