Suppose that [tex]$x=\ln(A)$[/tex] and [tex]$y=\ln(B)$[/tex]. Write the following expressions in terms of [tex][tex]$x$[/tex][/tex] and [tex]$y$[/tex] (so you cannot have an [tex]$A$[/tex] or [tex][tex]$B$[/tex][/tex] anywhere in your answers).

(a) [tex]\ln(AB) = [/tex]
(b) [tex]\ln\left(A^4 \cdot \sqrt{B}\right) = [/tex]
(c) [tex]\frac{\ln(A)}{\ln(B)} = [/tex]
(d) [tex]\ln\left(\frac{A}{B}\right) = [/tex]
(e) [tex]AB = [/tex]
(f) [tex]\ln(A-B) = [/tex]



Answer :

Alright, let's address each of the expressions step-by-step:

(a) [tex]\(\ln(AB)\)[/tex]:

Using the logarithm property [tex]\(\ln(AB) = \ln(A) + \ln(B)\)[/tex], we substitute the given values [tex]\(x = \ln(A)\)[/tex] and [tex]\(y = \ln(B)\)[/tex]. So, we get:
[tex]\[ \ln(AB) = \ln(A) + \ln(B) = x + y \][/tex]

(b) [tex]\(\ln\left(A^4 \cdot \sqrt{B}\right)\)[/tex]:

Using the properties of logarithms and exponents, we can break this down. First, [tex]\(A^4 \cdot \sqrt{B}\)[/tex] can be expressed as [tex]\(\ln(A^4) + \ln(\sqrt{B})\)[/tex]. Using the exponent rule [tex]\(\ln(A^n) = n \ln(A)\)[/tex], we have:
[tex]\[ \ln(A^4) = 4\ln(A) = 4x \][/tex]
Next, [tex]\(\sqrt{B}\)[/tex] is [tex]\(B^{1/2}\)[/tex], so:
[tex]\[ \ln(\sqrt{B}) = \ln(B^{1/2}) = (1/2)\ln(B) = \frac{1}{2}y \][/tex]
Combining these, we have:
[tex]\[ \ln\left(A^4 \cdot \sqrt{B}\right) = 4x + \frac{1}{2}y \][/tex]

(c) [tex]\(\frac{\ln(A)}{\ln(B)}\)[/tex]:

We are given [tex]\(x = \ln(A)\)[/tex] and [tex]\(y = \ln(B)\)[/tex]. So, directly substituting, we get:
[tex]\[ \frac{\ln(A)}{\ln(B)} = \frac{x}{y} \][/tex]

(d) [tex]\(\ln\left(\frac{A}{B}\right)\)[/tex]:

Using the logarithm property [tex]\(\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)\)[/tex], we substitute [tex]\(x = \ln(A)\)[/tex] and [tex]\(y = \ln(B)\)[/tex] to get:
[tex]\[ \ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B) = x - y \][/tex]

(e) [tex]\(AB\)[/tex]:

Expressing [tex]\(AB\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] using exponentials, since [tex]\(AB = e^{\ln(AB)}\)[/tex], and [tex]\(\ln(AB) = x + y\)[/tex], we get:
[tex]\[ AB = e^{\ln(A) + \ln(B)} = e^{x + y} \][/tex]

(f) [tex]\(\ln(A - B)\)[/tex]:

The expression [tex]\(\ln(A - B)\)[/tex] cannot be simplified directly in terms of [tex]\(\ln(A)\)[/tex] and [tex]\(\ln(B)\)[/tex] because logarithms do not have a simple property for the subtraction of numbers inside the logarithm. Therefore, [tex]\(\ln(A - B)\)[/tex] cannot be simplified using [tex]\(x\)[/tex] and [tex]\(y\)[/tex].

Recapping:

(a) [tex]\(\ln(AB) = x + y\)[/tex]

(b) [tex]\(\ln\left(A^4 \cdot \sqrt{B}\right) = 4x + (1/2)y\)[/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]\(AB = e^{x + y}\)[/tex]

(f) [tex]\(\ln(A - B)\)[/tex] cannot be simplified in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]