To rewrite the logarithmic expression [tex]\(\ln \left(\frac{e^3 r}{y^2}\right)\)[/tex] as the sum or difference of logs, we can use the properties of logarithms. Let's go through the steps:
1. Apply the logarithm of a quotient property:
The property states that [tex]\(\ln \left(\frac{A}{B}\right) = \ln A - \ln B\)[/tex]. Here, [tex]\(A = e^3 r\)[/tex] and [tex]\(B = y^2\)[/tex]. So,
[tex]\[
\ln \left(\frac{e^3 r}{y^2}\right) = \ln (e^3 r) - \ln (y^2).
\][/tex]
2. Apply the logarithm of a product property:
The property states that [tex]\(\ln (AB) = \ln A + \ln B\)[/tex]. For the term [tex]\(\ln (e^3 r)\)[/tex], we can split it into:
[tex]\[
\ln (e^3 r) = \ln (e^3) + \ln (r).
\][/tex]
3. Simplify the logarithm of an exponential:
The property states that [tex]\(\ln (e^k) = k\)[/tex]. Hence, for the term [tex]\(\ln (e^3)\)[/tex], we get:
[tex]\[
\ln (e^3) = 3.
\][/tex]
4. Combine all the results:
Now we can combine everything back together:
[tex]\[
\ln (e^3 r) - \ln (y^2) = [\ln (e^3) + \ln (r)] - \ln (y^2).
\][/tex]
We already simplified [tex]\(\ln (e^3)\)[/tex] to 3, so:
[tex]\[
\ln (e^3 r) - \ln (y^2) = [3 + \ln (r)] - \ln (y^2).
\][/tex]
5. Apply the logarithm of a power property:
The property states that [tex]\(\ln (A^k) = k \ln (A)\)[/tex]. For the term [tex]\(\ln (y^2)\)[/tex], we get:
[tex]\[
\ln (y^2) = 2 \ln (y).
\][/tex]
6. Final combination:
Substituting back, we get:
[tex]\[
\ln \left(\frac{e^3 r}{y^2}\right) = 3 + \ln (r) - 2 \ln (y).
\][/tex]
Hence, the logarithmic expression [tex]\(\ln \left(\frac{e^3 r}{y^2}\right)\)[/tex] can be written as:
[tex]\[
\ln \left(\frac{e^3 r}{y^2}\right) = \ln (r) - 2 \ln (y) + 3.
\][/tex]
