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Write the expression
(1)/(3)ln 27-3ln 5 as a single logarithm in simplest form without any negative exponents.
Answer:
ln(◻)

Write the expression $\frac{1}{3} \ln 27-3 \ln 5$ as a single logarithm in simplest form without any negative exponents. $\newline$ Answer: $\ln (\square)$

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Power rule with rational exponents

Full solution

Q. Write the expression

\frac{1}{3} \ln 27-3 \ln 5

as a single logarithm in simplest form without any negative exponents.

\newline

Answer:

\ln (\square)

Apply power rule: We have the expression $(\frac{1}{3})\ln 27 - 3\ln 5$ . We will use the logarithm power rule, which states that $a \cdot \ln(b) = \ln(b^a)$ , to simplify each term.
Calculate cube root: Apply the power rule to the first term: $(\frac{1}{3})\ln 27 = \ln(27^{\frac{1}{3}})$ . $\newline$ Calculate $27^{\frac{1}{3}}$ , which is the cube root of $27$ . $\newline$ $27^{\frac{1}{3}} = 3$ because $3^3 = 27$ . $\newline$ So, $(\frac{1}{3})\ln 27 = \ln(3)$ .
Apply power rule: Apply the power rule to the second term: $3\ln 5 = \ln(5^3)$ . Calculate $5^3$ . $5^3 = 125$ . So, $3\ln 5 = \ln(125)$ .
Combine logarithms: Now we have $\ln(3) - \ln(125)$ . We will use the logarithm quotient rule, which states that $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ , to combine the two logarithms into one.
Simplify to $\ln(\frac{3}{125})$ : Combine the logarithms: $\ln(3) - \ln(125) = \ln(\frac{3}{125})$ .
Simplify to $\ln(\frac{3}{125})$ : Combine the logarithms: $\ln(3) - \ln(125) = \ln(\frac{3}{125})$ .The expression is now simplified to a single logarithm: $\ln(\frac{3}{125})$ .

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