Answer :
Let's evaluate each expression step-by-step.
### Given:
[tex]\[ \ln a = 2, \quad \ln b = 3, \quad \ln c = 5 \][/tex]
### (a) Evaluate [tex]\(\ln \left(\frac{a^2}{b^2 c^3}\right)\)[/tex]
First, use the properties of logarithms to express the given term:
[tex]\[ \ln \left( \frac{a^2}{b^2 c^3} \right) = \ln(a^2) - \ln(b^2) - \ln(c^3) \][/tex]
Now, calculate each term individually:
[tex]\[ \ln(a^2) = 2 \cdot \ln(a) = 2 \cdot 2 = 4 \][/tex]
[tex]\[ \ln(b^2) = 2 \cdot \ln(b) = 2 \cdot 3 = 6 \][/tex]
[tex]\[ \ln(c^3) = 3 \cdot \ln(c) = 3 \cdot 5 = 15 \][/tex]
Combine these results:
[tex]\[ \ln \left( \frac{a^2}{b^2 c^3} \right) = 4 - 6 - 15 = -17 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{-17} \][/tex]
### (b) Evaluate [tex]\(\ln \sqrt{b^{-3} c a^{-1}}\)[/tex]
First, use the properties of logarithms to express the given term:
[tex]\[ \ln \sqrt{b^{-3} c a^{-1}} = \frac{1}{2} \ln(b^{-3} c a^{-1}) \][/tex]
Now, use logarithm properties again:
[tex]\[ \ln(b^{-3} c a^{-1}) = \ln(b^{-3}) + \ln(c) + \ln(a^{-1}) \][/tex]
[tex]\[ \ln(b^{-3}) = -3 \cdot \ln(b) = -3 \cdot 3 = -9 \][/tex]
[tex]\[ \ln(a^{-1}) = -1 \cdot \ln(a) = -1 \cdot 2 = -2 \][/tex]
Combine these results:
[tex]\[ \ln(b^{-3} c a^{-1}) = -9 + 5 - 2 = -6 \][/tex]
Then, evaluate:
[tex]\[ \ln \sqrt{b^{-3} c a^{-1}} = \frac{1}{2} \cdot (-6) = -3 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{-3} \][/tex]
### (c) Evaluate [tex]\(\frac{\ln \left(a^{-1} b^{-3}\right)}{\ln (b c)^{-2}}\)[/tex]
First, use the properties of logarithms to express the given term:
[tex]\[ \ln(a^{-1} b^{-3}) = \ln(a^{-1}) + \ln(b^{-3}) \][/tex]
[tex]\[ \ln(a^{-1}) = -\ln(a) = -2 \][/tex]
[tex]\[ \ln(b^{-3}) = -3 \cdot \ln(b) = -3 \cdot 3 = -9 \][/tex]
Combine these results:
[tex]\[ \ln(a^{-1} b^{-3}) = -2 - 9 = -11 \][/tex]
Next, consider the denominator:
[tex]\[ \ln(b c)^{-2} = -2 \cdot \ln(b c) \][/tex]
[tex]\[ \ln(b c) = \ln(b) + \ln(c) = 3 + 5 = 8 \][/tex]
[tex]\[ \ln(b c)^{-2} = -2 \cdot 8 = -16 \][/tex]
Combine the results to find the original expression:
[tex]\[ \frac{\ln(a^{-1} b^{-3})}{\ln(b c)^{-2}} = \frac{-11}{-16} = \frac{11}{16} \approx 0.6875 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{0.6875} \][/tex]
### (d) Evaluate [tex]\(\left(\ln c^4\right) \left( \ln \frac{a}{b^{-4}} \right)^{-2}\)[/tex]
First, use the properties of logarithms to express the given term:
[tex]\[ \ln(c^4) = 4 \cdot \ln(c) = 4 \cdot 5 = 20 \][/tex]
Next, consider:
[tex]\[ \ln \left(\frac{a}{b^{-4}}\right) = \ln(a) - \ln(b^{-4}) \][/tex]
[tex]\[ \ln(b^{-4}) = -4 \cdot \ln(b) = -4 \cdot 3 = -12 \][/tex]
Combine these results:
[tex]\[ \ln \frac{a}{b^{-4}} = \ln(a) + 12 = 2 + 12 = 14 \][/tex]
Finally, evaluate the given term:
[tex]\[ \left( \ln c^4 \right) \left( \ln \frac{a}{b^{-4}} \right)^{-2} = 20 \cdot \left( 14 \right)^{-2} = 20 \cdot \left( \frac{1}{196} \right) = \frac{20}{196} = \frac{10}{98} = \frac{5}{49} \approx 0.1020408163265306 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{0.1020408163265306} \][/tex]
These evaluations confirm the results step-by-step for each given expression.
### Given:
[tex]\[ \ln a = 2, \quad \ln b = 3, \quad \ln c = 5 \][/tex]
### (a) Evaluate [tex]\(\ln \left(\frac{a^2}{b^2 c^3}\right)\)[/tex]
First, use the properties of logarithms to express the given term:
[tex]\[ \ln \left( \frac{a^2}{b^2 c^3} \right) = \ln(a^2) - \ln(b^2) - \ln(c^3) \][/tex]
Now, calculate each term individually:
[tex]\[ \ln(a^2) = 2 \cdot \ln(a) = 2 \cdot 2 = 4 \][/tex]
[tex]\[ \ln(b^2) = 2 \cdot \ln(b) = 2 \cdot 3 = 6 \][/tex]
[tex]\[ \ln(c^3) = 3 \cdot \ln(c) = 3 \cdot 5 = 15 \][/tex]
Combine these results:
[tex]\[ \ln \left( \frac{a^2}{b^2 c^3} \right) = 4 - 6 - 15 = -17 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{-17} \][/tex]
### (b) Evaluate [tex]\(\ln \sqrt{b^{-3} c a^{-1}}\)[/tex]
First, use the properties of logarithms to express the given term:
[tex]\[ \ln \sqrt{b^{-3} c a^{-1}} = \frac{1}{2} \ln(b^{-3} c a^{-1}) \][/tex]
Now, use logarithm properties again:
[tex]\[ \ln(b^{-3} c a^{-1}) = \ln(b^{-3}) + \ln(c) + \ln(a^{-1}) \][/tex]
[tex]\[ \ln(b^{-3}) = -3 \cdot \ln(b) = -3 \cdot 3 = -9 \][/tex]
[tex]\[ \ln(a^{-1}) = -1 \cdot \ln(a) = -1 \cdot 2 = -2 \][/tex]
Combine these results:
[tex]\[ \ln(b^{-3} c a^{-1}) = -9 + 5 - 2 = -6 \][/tex]
Then, evaluate:
[tex]\[ \ln \sqrt{b^{-3} c a^{-1}} = \frac{1}{2} \cdot (-6) = -3 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{-3} \][/tex]
### (c) Evaluate [tex]\(\frac{\ln \left(a^{-1} b^{-3}\right)}{\ln (b c)^{-2}}\)[/tex]
First, use the properties of logarithms to express the given term:
[tex]\[ \ln(a^{-1} b^{-3}) = \ln(a^{-1}) + \ln(b^{-3}) \][/tex]
[tex]\[ \ln(a^{-1}) = -\ln(a) = -2 \][/tex]
[tex]\[ \ln(b^{-3}) = -3 \cdot \ln(b) = -3 \cdot 3 = -9 \][/tex]
Combine these results:
[tex]\[ \ln(a^{-1} b^{-3}) = -2 - 9 = -11 \][/tex]
Next, consider the denominator:
[tex]\[ \ln(b c)^{-2} = -2 \cdot \ln(b c) \][/tex]
[tex]\[ \ln(b c) = \ln(b) + \ln(c) = 3 + 5 = 8 \][/tex]
[tex]\[ \ln(b c)^{-2} = -2 \cdot 8 = -16 \][/tex]
Combine the results to find the original expression:
[tex]\[ \frac{\ln(a^{-1} b^{-3})}{\ln(b c)^{-2}} = \frac{-11}{-16} = \frac{11}{16} \approx 0.6875 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{0.6875} \][/tex]
### (d) Evaluate [tex]\(\left(\ln c^4\right) \left( \ln \frac{a}{b^{-4}} \right)^{-2}\)[/tex]
First, use the properties of logarithms to express the given term:
[tex]\[ \ln(c^4) = 4 \cdot \ln(c) = 4 \cdot 5 = 20 \][/tex]
Next, consider:
[tex]\[ \ln \left(\frac{a}{b^{-4}}\right) = \ln(a) - \ln(b^{-4}) \][/tex]
[tex]\[ \ln(b^{-4}) = -4 \cdot \ln(b) = -4 \cdot 3 = -12 \][/tex]
Combine these results:
[tex]\[ \ln \frac{a}{b^{-4}} = \ln(a) + 12 = 2 + 12 = 14 \][/tex]
Finally, evaluate the given term:
[tex]\[ \left( \ln c^4 \right) \left( \ln \frac{a}{b^{-4}} \right)^{-2} = 20 \cdot \left( 14 \right)^{-2} = 20 \cdot \left( \frac{1}{196} \right) = \frac{20}{196} = \frac{10}{98} = \frac{5}{49} \approx 0.1020408163265306 \][/tex]
Thus, the answer is:
[tex]\[ \boxed{0.1020408163265306} \][/tex]
These evaluations confirm the results step-by-step for each given expression.
