Answer :
Certainly! Let's address how each logarithm can be expressed in terms of [tex]\(\ln 3\)[/tex] and [tex]\(\ln 5\)[/tex].
### 1. [tex]\(\ln \frac{81}{125}\)[/tex]
We begin by simplifying [tex]\(\ln \frac{81}{125}\)[/tex].
Using properties of logarithms:
[tex]\[ \ln \frac{a}{b} = \ln a - \ln b \][/tex]
we can rewrite:
[tex]\[ \ln \frac{81}{125} = \ln 81 - \ln 125 \][/tex]
Next, we express [tex]\(\ln 81\)[/tex] and [tex]\(\ln 125\)[/tex] in terms of [tex]\(\ln 3\)[/tex] and [tex]\(\ln 5\)[/tex].
[tex]\[ 81 = 3^4 \implies \ln 81 = \ln (3^4) = 4 \ln 3 \][/tex]
[tex]\[ 125 = 5^3 \implies \ln 125 = \ln (5^3) = 3 \ln 5 \][/tex]
Now, substituting these back into the equation, we get:
[tex]\[ \ln \frac{81}{125} = 4 \ln 3 - 3 \ln 5 \][/tex]
Thus, [tex]\(\ln \frac{81}{125}\)[/tex] is expressed in terms of [tex]\(\ln 3\)[/tex] and [tex]\(\ln 5\)[/tex] as:
[tex]\[ \boxed{4 \ln 3 - 3 \ln 5} \][/tex]
### 2. [tex]\(4 \ln 5 - 3 \ln 3\)[/tex]
Since this is already expressed in terms of [tex]\(\ln 3\)[/tex] and [tex]\(\ln 5\)[/tex], we do not need to perform any additional calculations:
[tex]\[ \boxed{4 \ln 5 - 3 \ln 3} \][/tex]
### 3. [tex]\(5 \ln 3 - 3 \ln 4\)[/tex]
We need to express [tex]\(\ln 4\)[/tex] in terms of [tex]\(\ln 5\)[/tex] or [tex]\(\ln 3\)[/tex].
Knowing that:
[tex]\[ 4 = 2^2 \implies \ln 4 = \ln (2^2) = 2 \ln 2 \][/tex]
But, [tex]\(\ln 2\)[/tex] is not directly related to [tex]\(\ln 3\)[/tex] or [tex]\(\ln 5\)[/tex] in a simple form, so we will leave this expression in terms of [tex]\(\ln 4\)[/tex].
Thus:
[tex]\[ \boxed{5 \ln 3 - 3 \ln 4} \][/tex]
### 4. [tex]\(4 \ln 3 - 3 \ln 5\)[/tex]
This is exactly the same as what we derived for [tex]\(\ln \frac{81}{125}\)[/tex]:
[tex]\[ \boxed{4 \ln 3 - 3 \ln 5} \][/tex]
### 5. [tex]\(3 \ln 4 - 5 \ln 3\)[/tex]
Similarly to before, [tex]\(\ln 4\)[/tex] is related to [tex]\(\ln 2\)[/tex], but for simplicity, we leave the expression:
[tex]\[ \boxed{3 \ln 4 - 5 \ln 3} \][/tex]
In conclusion:
- [tex]\(\ln \frac{81}{125} = 4 \ln 3 - 3 \ln 5\)[/tex]
- [tex]\(4 \ln 5 - 3 \ln 3\)[/tex]
- [tex]\(5 \ln 3 - 3 \ln 4\)[/tex]
- [tex]\(4 \ln 3 - 3 \ln 5\)[/tex]
- [tex]\(3 \ln 4 - 5 \ln 3\)[/tex]
So,
- The first expression [tex]\(\ln \frac{81}{125}\)[/tex] simplifies to [tex]\(4 \ln 3 - 3 \ln 5\)[/tex].
- The second expression [tex]\(4 \ln 5 - 3 \ln 3\)[/tex] is given as is.
- The third expression [tex]\(5 \ln 3 - 3 \ln 4\)[/tex] is given as is and includes [tex]\(\ln 4\)[/tex].
- The fourth expression [tex]\(4 \ln 3 - 3 \ln 5\)[/tex] matches what we derived above for [tex]\(\ln 81/125\)[/tex].
- The fifth expression [tex]\(3 \ln 4 - 5 \ln 3\)[/tex] is in terms of [tex]\(\ln 4\)[/tex].
Hence, the equivalent answer to the given logarithmic expressions confirms that [tex]\( \ln \frac{81}{125} = 4 \ln 3 - 3 \ln 5 \)[/tex].
### 1. [tex]\(\ln \frac{81}{125}\)[/tex]
We begin by simplifying [tex]\(\ln \frac{81}{125}\)[/tex].
Using properties of logarithms:
[tex]\[ \ln \frac{a}{b} = \ln a - \ln b \][/tex]
we can rewrite:
[tex]\[ \ln \frac{81}{125} = \ln 81 - \ln 125 \][/tex]
Next, we express [tex]\(\ln 81\)[/tex] and [tex]\(\ln 125\)[/tex] in terms of [tex]\(\ln 3\)[/tex] and [tex]\(\ln 5\)[/tex].
[tex]\[ 81 = 3^4 \implies \ln 81 = \ln (3^4) = 4 \ln 3 \][/tex]
[tex]\[ 125 = 5^3 \implies \ln 125 = \ln (5^3) = 3 \ln 5 \][/tex]
Now, substituting these back into the equation, we get:
[tex]\[ \ln \frac{81}{125} = 4 \ln 3 - 3 \ln 5 \][/tex]
Thus, [tex]\(\ln \frac{81}{125}\)[/tex] is expressed in terms of [tex]\(\ln 3\)[/tex] and [tex]\(\ln 5\)[/tex] as:
[tex]\[ \boxed{4 \ln 3 - 3 \ln 5} \][/tex]
### 2. [tex]\(4 \ln 5 - 3 \ln 3\)[/tex]
Since this is already expressed in terms of [tex]\(\ln 3\)[/tex] and [tex]\(\ln 5\)[/tex], we do not need to perform any additional calculations:
[tex]\[ \boxed{4 \ln 5 - 3 \ln 3} \][/tex]
### 3. [tex]\(5 \ln 3 - 3 \ln 4\)[/tex]
We need to express [tex]\(\ln 4\)[/tex] in terms of [tex]\(\ln 5\)[/tex] or [tex]\(\ln 3\)[/tex].
Knowing that:
[tex]\[ 4 = 2^2 \implies \ln 4 = \ln (2^2) = 2 \ln 2 \][/tex]
But, [tex]\(\ln 2\)[/tex] is not directly related to [tex]\(\ln 3\)[/tex] or [tex]\(\ln 5\)[/tex] in a simple form, so we will leave this expression in terms of [tex]\(\ln 4\)[/tex].
Thus:
[tex]\[ \boxed{5 \ln 3 - 3 \ln 4} \][/tex]
### 4. [tex]\(4 \ln 3 - 3 \ln 5\)[/tex]
This is exactly the same as what we derived for [tex]\(\ln \frac{81}{125}\)[/tex]:
[tex]\[ \boxed{4 \ln 3 - 3 \ln 5} \][/tex]
### 5. [tex]\(3 \ln 4 - 5 \ln 3\)[/tex]
Similarly to before, [tex]\(\ln 4\)[/tex] is related to [tex]\(\ln 2\)[/tex], but for simplicity, we leave the expression:
[tex]\[ \boxed{3 \ln 4 - 5 \ln 3} \][/tex]
In conclusion:
- [tex]\(\ln \frac{81}{125} = 4 \ln 3 - 3 \ln 5\)[/tex]
- [tex]\(4 \ln 5 - 3 \ln 3\)[/tex]
- [tex]\(5 \ln 3 - 3 \ln 4\)[/tex]
- [tex]\(4 \ln 3 - 3 \ln 5\)[/tex]
- [tex]\(3 \ln 4 - 5 \ln 3\)[/tex]
So,
- The first expression [tex]\(\ln \frac{81}{125}\)[/tex] simplifies to [tex]\(4 \ln 3 - 3 \ln 5\)[/tex].
- The second expression [tex]\(4 \ln 5 - 3 \ln 3\)[/tex] is given as is.
- The third expression [tex]\(5 \ln 3 - 3 \ln 4\)[/tex] is given as is and includes [tex]\(\ln 4\)[/tex].
- The fourth expression [tex]\(4 \ln 3 - 3 \ln 5\)[/tex] matches what we derived above for [tex]\(\ln 81/125\)[/tex].
- The fifth expression [tex]\(3 \ln 4 - 5 \ln 3\)[/tex] is in terms of [tex]\(\ln 4\)[/tex].
Hence, the equivalent answer to the given logarithmic expressions confirms that [tex]\( \ln \frac{81}{125} = 4 \ln 3 - 3 \ln 5 \)[/tex].
