Sure! Let's solve each part step-by-step using properties of logarithms.
### Part (a)
Evaluate [tex]\(\log_5 45 - 2 \log_5 3\)[/tex]:
1. Use the property of logarithms which states [tex]\(a \log_b(x) = \log_b(x^a)\)[/tex]:
[tex]\[
2 \log_5 3 = \log_5 (3^2)
\][/tex]
Simplify [tex]\(3^2\)[/tex]:
[tex]\[
\log_5 9
\][/tex]
2. Substitute [tex]\(\log_5 9\)[/tex] back into the original expression:
[tex]\[
\log_5 45 - \log_5 9
\][/tex]
3. Use the logarithmic property [tex]\(\log_b(x) - \log_b(y) = \log_b\left(\frac{x}{y}\right)\)[/tex]:
[tex]\[
\log_5 \left(\frac{45}{9}\right)
\][/tex]
4. Simplify [tex]\(\frac{45}{9}\)[/tex]:
[tex]\[
\log_5 5
\][/tex]
5. Since [tex]\(\log_5 5\)[/tex] is the logarithm of 5 base 5, it simplifies to:
[tex]\[
1
\][/tex]
So the solution to (a) is:
[tex]\[
\log_5 45 - 2 \log_5 3 = 1
\][/tex]
### Part (b)
Evaluate [tex]\(\ln e^{-9} + \ln e^5\)[/tex]:
1. Use the property [tex]\(\ln(e^x) = x\)[/tex]:
[tex]\[
\ln e^{-9} = -9 \quad \text{and} \quad \ln e^5 = 5
\][/tex]
2. Substitute these values back into the expression:
[tex]\[
-9 + 5
\][/tex]
3. Perform the addition:
[tex]\[
-4
\][/tex]
So the solution to (b) is:
[tex]\[
\ln e^{-9} + \ln e^5 = -4
\][/tex]
Combining the results, we have:
(a) [tex]\(\log_5 45 - 2 \log_5 3 = 1\)[/tex]
(b) [tex]\(\ln e^{-9} + \ln e^5 = -4\)[/tex]
