Given the substitutions ln 2=a,ln 3=b, and ln 5=c, find the value of ln((16)/(27)) in terms of a,b, and c. Answer:

Rewrite in Prime Factors: Given that ln 2 = a, ln 3 = b, and ln 5 = c, we need to express ln((16)/(27)) using these substitutions. First, we can rewrite 16 and 27 in terms of their prime factors: 16 = 2^4 and 27 = 3^3. So, ln((16)/(27)) becomes ln((2^4)/(3^3)). Using the properties of logarithms, we can express this as ln(2^4) - ln(3^3).Apply Power Rule: Now, we apply the power rule of logarithms, which states that ln(x^n) = n * ln(x), to both terms. This gives us 4 * ln(2) - 3 * ln(3).Substitute Given Values: Substitute the given values for ln(2) and ln(3), which are a and b respectively. This results in 4 * a - 3 * b.Final Expression: We have now expressed ln((16)/(27)) in terms of a and b. Since ln(5) = c is not needed to express ln((16)/(27)), we do not use it in our final expression.

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Given the substitutions
ln 2=a,ln 3=b, and
ln 5=c, find the value of
ln((16)/(27)) 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 $\ln \left(\frac{16}{27}\right)$ in terms of $a, b$ , and $c$ . $\newline$ Answer:

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Algebra 1

Write and solve inverse variation equations

Full solution

Q. Given the substitutions

\ln 2=a, \ln 3=b

, and

\ln 5=c

, find the value of

\ln \left(\frac{16}{27}\right)

in terms of

a, b

, and

c

\newline

Answer:

Rewrite in Prime Factors: Given that $\ln 2 = a$ , $\ln 3 = b$ , and $\ln 5 = c$ , we need to express $\ln\left(\frac{16}{27}\right)$ using these substitutions. $\newline$ First, we can rewrite $16$ and $27$ in terms of their prime factors: $16 = 2^4$ and $27 = 3^3$ . $\newline$ So, $\ln\left(\frac{16}{27}\right)$ becomes $\ln\left(\frac{2^4}{3^3}\right)$ . $\newline$ Using the properties of logarithms, we can express this as $\ln 3 = b$ $0$ .
Apply Power Rule: Now, we apply the power rule of logarithms, which states that $\ln(x^n) = n \times \ln(x)$ , to both terms. $\newline$ This gives us $4 \times \ln(2) - 3 \times \ln(3)$ .
Substitute Given Values: Substitute the given values for $\ln(2)$ and $\ln(3)$ , which are $a$ and $b$ respectively. $\newline$ This results in $4 \times a - 3 \times b$ .
Final Expression: We have now expressed $\ln\left(\frac{16}{27}\right)$ in terms of $a$ and $b$ . Since $\ln(5) = c$ is not needed to express $\ln\left(\frac{16}{27}\right)$ , we do not use it in our final expression.