Answer :
Sure! Let's work through each expression step-by-step and convert them into terms of [tex]\( x \)[/tex] and [tex]\( y \)[/tex], ensuring we clearly understand how the logarithmic properties are applied.
Given [tex]\( x = \ln(A) \)[/tex] and [tex]\( y = \ln(B) \)[/tex]:
### (a) [tex]\(\ln(AB)\)[/tex]
Using the property of logarithms that [tex]\(\ln(AB) = \ln(A) + \ln(B)\)[/tex], we have:
[tex]\[ \ln(AB) = \ln(A) + \ln(B) = x + y \][/tex]
### (b) [tex]\(\ln \left(A^2 \cdot \sqrt{B}\right)\)[/tex]
First, we rewrite [tex]\( \sqrt{B} \)[/tex] as [tex]\( B^{1/2} \)[/tex]. Then, using the logarithm property [tex]\(\ln(xy) = \ln(x) + \ln(y)\)[/tex], we obtain:
[tex]\[ \ln \left(A^2 \cdot \sqrt{B}\right) = \ln(A^2) + \ln(B^{1/2}) \][/tex]
Next, apply the property that [tex]\(\ln(x^n) = n \ln(x)\)[/tex]:
[tex]\[ \ln(A^2) = 2 \ln(A) = 2x \][/tex]
[tex]\[ \ln(B^{1/2}) = \frac{1}{2} \ln(B) = \frac{1}{2} y \][/tex]
Combining these:
[tex]\[ \ln \left(A^2 \cdot \sqrt{B}\right) = 2x + \frac{1}{2} y \][/tex]
### (c) [tex]\(\frac{\ln(A)}{\ln(B)}\)[/tex]
This can be directly written as:
[tex]\[ \frac{\ln(A)}{\ln(B)} = \frac{x}{y} \][/tex]
### (d) [tex]\(\ln \left(\frac{A}{B}\right)\)[/tex]
Using the logarithm property that [tex]\(\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)\)[/tex]:
[tex]\[ \ln \left(\frac{A}{B}\right) = \ln(A) - \ln(B) = x - y \][/tex]
### (e) [tex]\(AB\)[/tex]
Using the exponential property of logarithms that if [tex]\(\ln(A) = x\)[/tex], then [tex]\(A = e^x\)[/tex]:
[tex]\[ A = e^x \quad \text{and} \quad B = e^y \][/tex]
Hence:
[tex]\[ AB = e^x \cdot e^y = e^{x+y} \][/tex]
To convert this into base 10, we recognize [tex]\(e\)[/tex] can be converted to 10 using the property [tex]\(e^k = 10^{k \ln(10)}\)[/tex]. Thus,
[tex]\[ AB = 10^{(x+y)/\ln(10)} \][/tex]
However, using [tex]\(e\)[/tex]:
[tex]\[ AB = 10^{x+y} \][/tex]
### (f) [tex]\(\ln(A - B)\)[/tex]
Unfortunately, logarithms do not have a simple property that allows [tex]\(\ln(A - B)\)[/tex] to be written in terms of [tex]\(\ln(A)\)[/tex] and [tex]\(\ln(B)\)[/tex]. Logarithmic identities mainly deal with multiplication, division, and powers, not subtraction.
Thus,
[tex]\[ \ln (A - B) \quad \text{cannot be simplified using just } x \text{ and } y. \][/tex]
To summarize:
(a) [tex]\(\ln(AB) = x + y\)[/tex]
(b) [tex]\(\ln \left(A^2 \cdot \sqrt{B}\right) = 2x + \frac{1}{2} y\)[/tex]
(c) [tex]\(\frac{\ln(A)}{\ln(B)} = \frac{x}{y}\)[/tex]
(d) [tex]\(\ln \left(\frac{A}{B}\right) = x - y\)[/tex]
(e) [tex]\(AB = 10^{x+y}\)[/tex]
(f) [tex]\(\ln (A - B) \)[/tex] does not have a simplifiable form in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
Given [tex]\( x = \ln(A) \)[/tex] and [tex]\( y = \ln(B) \)[/tex]:
### (a) [tex]\(\ln(AB)\)[/tex]
Using the property of logarithms that [tex]\(\ln(AB) = \ln(A) + \ln(B)\)[/tex], we have:
[tex]\[ \ln(AB) = \ln(A) + \ln(B) = x + y \][/tex]
### (b) [tex]\(\ln \left(A^2 \cdot \sqrt{B}\right)\)[/tex]
First, we rewrite [tex]\( \sqrt{B} \)[/tex] as [tex]\( B^{1/2} \)[/tex]. Then, using the logarithm property [tex]\(\ln(xy) = \ln(x) + \ln(y)\)[/tex], we obtain:
[tex]\[ \ln \left(A^2 \cdot \sqrt{B}\right) = \ln(A^2) + \ln(B^{1/2}) \][/tex]
Next, apply the property that [tex]\(\ln(x^n) = n \ln(x)\)[/tex]:
[tex]\[ \ln(A^2) = 2 \ln(A) = 2x \][/tex]
[tex]\[ \ln(B^{1/2}) = \frac{1}{2} \ln(B) = \frac{1}{2} y \][/tex]
Combining these:
[tex]\[ \ln \left(A^2 \cdot \sqrt{B}\right) = 2x + \frac{1}{2} y \][/tex]
### (c) [tex]\(\frac{\ln(A)}{\ln(B)}\)[/tex]
This can be directly written as:
[tex]\[ \frac{\ln(A)}{\ln(B)} = \frac{x}{y} \][/tex]
### (d) [tex]\(\ln \left(\frac{A}{B}\right)\)[/tex]
Using the logarithm property that [tex]\(\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)\)[/tex]:
[tex]\[ \ln \left(\frac{A}{B}\right) = \ln(A) - \ln(B) = x - y \][/tex]
### (e) [tex]\(AB\)[/tex]
Using the exponential property of logarithms that if [tex]\(\ln(A) = x\)[/tex], then [tex]\(A = e^x\)[/tex]:
[tex]\[ A = e^x \quad \text{and} \quad B = e^y \][/tex]
Hence:
[tex]\[ AB = e^x \cdot e^y = e^{x+y} \][/tex]
To convert this into base 10, we recognize [tex]\(e\)[/tex] can be converted to 10 using the property [tex]\(e^k = 10^{k \ln(10)}\)[/tex]. Thus,
[tex]\[ AB = 10^{(x+y)/\ln(10)} \][/tex]
However, using [tex]\(e\)[/tex]:
[tex]\[ AB = 10^{x+y} \][/tex]
### (f) [tex]\(\ln(A - B)\)[/tex]
Unfortunately, logarithms do not have a simple property that allows [tex]\(\ln(A - B)\)[/tex] to be written in terms of [tex]\(\ln(A)\)[/tex] and [tex]\(\ln(B)\)[/tex]. Logarithmic identities mainly deal with multiplication, division, and powers, not subtraction.
Thus,
[tex]\[ \ln (A - B) \quad \text{cannot be simplified using just } x \text{ and } y. \][/tex]
To summarize:
(a) [tex]\(\ln(AB) = x + y\)[/tex]
(b) [tex]\(\ln \left(A^2 \cdot \sqrt{B}\right) = 2x + \frac{1}{2} y\)[/tex]
(c) [tex]\(\frac{\ln(A)}{\ln(B)} = \frac{x}{y}\)[/tex]
(d) [tex]\(\ln \left(\frac{A}{B}\right) = x - y\)[/tex]
(e) [tex]\(AB = 10^{x+y}\)[/tex]
(f) [tex]\(\ln (A - B) \)[/tex] does not have a simplifiable form in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
