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Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log x,log y, and
log z.

log ((x^(5)y^(4))/(root(3)(z^(4))))
Answer:

Expand the logarithm fully using the properties of logs. Express the final answer in terms of $\log x, \log y$ , and $\log z$ . $\newline$ $\log \frac{x^{5} y^{4}}{\sqrt[3]{z^{4}}}$ $\newline$ Answer:

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Evaluate definite integrals using the power rule

Full solution

Q. Expand the logarithm fully using the properties of logs. Express the final answer in terms of

\log x, \log y

, and

\log z

.

\newline

\log \frac{x^{5} y^{4}}{\sqrt[3]{z^{4}}}

\newline

Answer:

Use Quotient Rule: Use the quotient rule for logarithms, which states that $\log\left(\frac{a}{b}\right) = \log(a) - \log(b)$ , to separate the numerator and the denominator. $\newline$ $\log\left(\frac{x^{5}y^{4}}{\sqrt[3]{z^{4}}}\right) = \log(x^{5}y^{4}) - \log(\sqrt[3]{z^{4}})$
Apply Product Rule: Apply the product rule for logarithms, which states that $\log(ab) = \log(a) + \log(b)$ , to the numerator. $\newline$ $\log(x^{5}y^{4}) = \log(x^{5}) + \log(y^{4})$
Use Power Rule: Use the power rule for logarithms, which states that $\log(a^n) = n\log(a)$ , to take the exponents out in front of the logs. $\newline$ $\log(x^{5}) + \log(y^{4}) = 5\log(x) + 4\log(y)$
Apply Power Rule: Apply the power rule for logarithms to the denominator, noting that the cube root of $z^4$ is $z^{(4/3)}$ . $\newline$ $\log(\sqrt[3]{z^{4}}) = \log(z^{4/3})$
Combine Results: Again, use the power rule for logarithms to bring the exponent out in front of the log in the denominator. $\newline$ $\log(z^{\frac{4}{3}}) = \frac{4}{3}\cdot\log(z)$
Combine Results: Again, use the power rule for logarithms to bring the exponent out in front of the log in the denominator. $\newline$ $\log(z^{\frac{4}{3}}) = \frac{4}{3}\log(z)$ Combine the results from the numerator and the denominator using the result from the first step. $\newline$ $\log\left(\frac{x^{5}y^{4}}{\sqrt[3]{z^{4}}}\right) = (5\log(x) + 4\log(y)) - \frac{4}{3}\log(z)$

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