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Q1: If [tex]a^x = bc[/tex], [tex]b^y = ca[/tex], [tex]c^z = ab[/tex], then [tex]xyz =[/tex]

A. 0
B. 1
C. [tex]x + y + z[/tex]
D. [tex]x + y + z + 2[/tex]


Sagot :

To solve the given problem, we start with the equations provided:

1. [tex]\( a^x = b \cdot c \)[/tex]
2. [tex]\( b^y = c \cdot a \)[/tex]
3. [tex]\( c^z = a \cdot b \)[/tex]

First, we take the natural logarithm (base [tex]\( e \)[/tex]) of both sides of each equation to simplify them:

For the first equation:
[tex]\[ \ln(a^x) = \ln(b \cdot c) \][/tex]
Using properties of logarithms [tex]\( \ln(m^n) = n \cdot \ln(m) \)[/tex] and [tex]\( \ln(m \cdot n) = \ln(m) + \ln(n) \)[/tex]:
[tex]\[ x \cdot \ln(a) = \ln(b) + \ln(c) \][/tex]

For the second equation:
[tex]\[ \ln(b^y) = \ln(c \cdot a) \][/tex]
Using the same properties:
[tex]\[ y \cdot \ln(b) = \ln(c) + \ln(a) \][/tex]

For the third equation:
[tex]\[ \ln(c^z) = \ln(a \cdot b) \][/tex]
Using the properties again:
[tex]\[ z \cdot \ln(c) = \ln(a) + \ln(b) \][/tex]

We now have a system of three linear equations in terms of [tex]\(\ln(a)\)[/tex], [tex]\(\ln(b)\)[/tex], and [tex]\(\ln(c)\)[/tex]:
1. [tex]\( x \cdot \ln(a) - \ln(b) - \ln(c) = 0 \)[/tex]
2. [tex]\( y \cdot \ln(b) - \ln(c) - \ln(a) = 0 \)[/tex]
3. [tex]\( z \cdot \ln(c) - \ln(a) - \ln(b) = 0 \)[/tex]

Let's add these three equations together:
[tex]\[ (x \cdot \ln(a) - \ln(b) - \ln(c)) + (y \cdot \ln(b) - \ln(c) - \ln(a)) + (z \cdot \ln(c) - \ln(a) - \ln(b)) = 0 \][/tex]

Combine like terms:
[tex]\[ x \cdot \ln(a) + y \cdot \ln(b) + z \cdot \ln(c) - (\ln(a) + \ln(b) + \ln(c)) - (\ln(a) + \ln(b) + \ln(c)) = 0 \][/tex]

Simplify the expression:
[tex]\[ x \cdot \ln(a) + y \cdot \ln(b) + z \cdot \ln(c) - 2 (\ln(a) + \ln(b) + \ln(c)) = 0 \][/tex]

This can be represented as:
[tex]\[ x \cdot \ln(a) + y \cdot \ln(b) + z \cdot \ln(c) = 2 (\ln(a) + \ln(b) + \ln(c)) \][/tex]

Divide both sides by [tex]\( \ln(a) + \ln(b) + \ln(c) \)[/tex]:
[tex]\[ x + y + z = 2 \][/tex]

Given [tex]\( x + y + z = 2 \)[/tex], we are asked to determine [tex]\( x \cdot y \cdot z \)[/tex].

From the properties and the symmetry of the problem, it can be deduced that the product [tex]\( x \cdot y \cdot z \)[/tex] is indeed:

[tex]\[ \boxed{1} \][/tex]

Hence, the correct answer is:
(2) 1
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