Answer :
Let's solve the given problems step-by-step. We will express each expression as a single logarithm using logarithmic properties.
a. Express [tex]2\log_{10} a - \frac{1}{3}\log_{10} b + 2[/tex] as a single logarithm.
To express this as a single logarithm, we'll use the properties of logarithms:
- Power Rule: [tex]k\log_b M = \log_b M^k[/tex]
- Product Rule: [tex]\log_b M + \log_b N = \log_b (MN)[/tex]
- Quotient Rule: [tex]\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)[/tex]
Let's apply these properties step-by-step:
Apply the Power Rule to each term:
- [tex]2\log_{10} a = \log_{10} (a^2)[/tex]
- [tex]\frac{1}{3}\log_{10} b = \log_{10} (b^{1/3})[/tex]
This makes the expression:
[tex]\log_{10} (a^2) - \log_{10} (b^{1/3}) + 2[/tex]The constant 2 needs to be expressed in terms of a logarithm base 10. We note that:
[tex]2 = \log_{10}(10^2) = \log_{10} (100)[/tex]Combine all terms using the Product and Quotient Rules:
[tex]\log_{10} \left(\frac{a^2 \cdot 100}{b^{1/3}}\right)[/tex]
Thus, the expression as a single logarithm is:
[tex]\log_{10} \left(\frac{100a^2}{b^{1/3}}\right)[/tex]
b. Express [tex]\log_c(x+1) - \log_c(x-1) + A[/tex] as a single logarithm, where [tex]A = \log_c B[/tex].
Let's follow the same logarithmic rules to simplify this expression:
Substitute [tex]A = \log_c B[/tex] into the expression:
[tex]\log_c(x+1) - \log_c(x-1) + \log_c B[/tex]Apply the Quotient Rule to [tex]\log_c(x+1) - \log_c(x-1)[/tex]:
[tex]\log_c \left(\frac{x+1}{x-1}\right)[/tex]Combine with [tex]\log_c B[/tex] using the Product Rule:
[tex]\log_c \left(\frac{x+1}{x-1} \cdot B\right)[/tex]
Thus, the expression as a single logarithm is:
[tex]\log_c \left(\frac{B(x+1)}{x-1}\right)[/tex]