Given the substitutions ln 2=a,ln 3=b, and ln 5=c, find the value of ln((root(3)(5))/(16)) in terms of a,b, and c. Answer:

Break down logarithm: We need to express ln((root(3)(5))/(16)) using the given substitutions ln 2=a, ln 3=b, and ln 5=c. We start by breaking down the logarithm using logarithmic properties. ln((root(3)(5))/(16)) = ln(root(3)(5)) - ln(16)Express in prime factors: Next, we express the cube root and the number 16 in terms of their prime factors to simplify the logarithms. ln(root(3)(5)) can be written as (1/3)ln(5) because the cube root is the same as raising to the power of 1/3. ln(16) can be written as ln(2^4) because 16 is 2 raised to the power of 4.Apply power rule: Now we apply the power rule of logarithms, which states that ln(x^y) = y*ln(x), to both terms. (1/3)ln(5) becomes (1/3)c because ln 5=c. ln(2^4) becomes 4*ln(2) because ln(2)=a.Substitute values: Substitute the values of a, b, and c into the expression. ln((root(3)(5))/(16)) = (1/3)c - 4aFinal expression: We have now expressed ln((root(3)(5))/(16)) in terms of a, b, and c. There are no further simplifications needed, and we have not made any mathematical errors.

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Given the substitutions
ln 2=a,ln 3=b, and
ln 5=c, find the value of
ln((root(3)(5))/(16)) in terms of
a,b, and
c.
Answer:

Given the substitutions $\ln 2=a, \ln 3=b$ , and $\ln 5=c$ , find the value of $\ln \left(\frac{\sqrt[3]{5}}{16}\right)$ in terms of $a, b$ , and $c$ . $\newline$ Answer:

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Q. Given the substitutions

\ln 2=a, \ln 3=b

, and

\ln 5=c

, find the value of

\ln \left(\frac{\sqrt[3]{5}}{16}\right)

in terms of

a, b

, and

c

\newline

Answer:

Break down logarithm: We need to express $\ln\left(\frac{\sqrt[3]{5}}{16}\right)$ using the given substitutions $\ln 2=a$ , $\ln 3=b$ , and $\ln 5=c$ . We start by breaking down the logarithm using logarithmic properties. $\newline$ $\ln\left(\frac{\sqrt[3]{5}}{16}\right) = \ln(\sqrt[3]{5}) - \ln(16)$
Express in prime factors: Next, we express the cube root and the number $16$ in terms of their prime factors to simplify the logarithms. $\newline$ $\ln(\sqrt[3]{5})$ can be written as $(\frac{1}{3})\ln(5)$ because the cube root is the same as raising to the power of $\frac{1}{3}$ . $\newline$ $\ln(16)$ can be written as $\ln(2^4)$ because $16$ is $2$ raised to the power of $4$ .
Apply power rule: Now we apply the power rule of logarithms, which states that $\ln(x^y) = y\cdot\ln(x)$ , to both terms. $\newline$ $(1/3)\ln(5)$ becomes $(1/3)c$ because $\ln 5=c$ . $\newline$ $\ln(2^4)$ becomes $4\cdot\ln(2)$ because $\ln(2)=a$ .
Substitute values: Substitute the values of $a$ , $b$ , and $c$ into the expression. $\ln\left(\frac{\sqrt[3]{5}}{16}\right) = \frac{1}{3}c - 4a$
Final expression: We have now expressed $\ln\left(\frac{\sqrt[3]{5}}{16}\right)$ in terms of $a$ , $b$ , and $c$ . There are no further simplifications needed, and we have not made any mathematical errors.