Sure, let's solve the problem step-by-step and prove that [tex]\(\frac{3}{b} = \frac{1}{a} + \frac{1}{c}\)[/tex].
We are given two conditions:
1. [tex]\(x^a = y^b = z^c\)[/tex]
2. [tex]\(y^3 = xz\)[/tex]
To deal with these, we will employ logarithms and algebraic manipulation.
### Step 1: Expression of each quantity in terms of a common variable
Let's assume a common variable [tex]\(k\)[/tex] such that:
[tex]\[ x^a = y^b = z^c = k \][/tex]
Taking the logarithm of both sides for each equation:
[tex]\[ \log(x^a) = \log(k) \][/tex]
[tex]\[ \log(y^b) = \log(k) \][/tex]
[tex]\[ \log(z^c) = \log(k) \][/tex]
Using the properties of logarithms (specifically, [tex]\(\log(m^n) = n \log(m)\)[/tex]):
[tex]\[ a \log(x) = \log(k) \][/tex]
[tex]\[ b \log(y) = \log(k) \][/tex]
[tex]\[ c \log(z) = \log(k) \][/tex]
From these equations, solve for [tex]\(\log(x)\)[/tex], [tex]\(\log(y)\)[/tex], and [tex]\(\log(z)\)[/tex]:
[tex]\[ \log(x) = \frac{\log(k)}{a} \][/tex]
[tex]\[ \log(y) = \frac{\log(k)}{b} \][/tex]
[tex]\[ \log(z) = \frac{\log(k)}{c} \][/tex]
### Step 2: Substitute into the second condition
Given the second condition [tex]\(y^3 = xz\)[/tex], we take the logarithm of both sides:
[tex]\[ \log(y^3) = \log(xz) \][/tex]
Using properties of logarithms:
[tex]\[ 3 \log(y) = \log(x) + \log(z) \][/tex]
### Step 3: Substitute the expressions for [tex]\(\log(x)\)[/tex], [tex]\(\log(y)\)[/tex], and [tex]\(\log(z)\)[/tex]
[tex]\[ 3 \left(\frac{\log(k)}{b} \right) = \left(\frac{\log(k)}{a}\right) + \left(\frac{\log(k)}{c}\right) \][/tex]
Factor out [tex]\(\log(k)\)[/tex] from all terms:
[tex]\[ 3 \frac{\log(k)}{b} = \frac{\log(k)}{a} + \frac{\log(k)}{c} \][/tex]
If [tex]\(\log(k) \neq 0\)[/tex] (since [tex]\(k > 0\)[/tex]):
[tex]\[ 3 \frac{1}{b} = \frac{1}{a} + \frac{1}{c} \][/tex]
This simplifies exactly to the statement we wanted to prove:
[tex]\[ \frac{3}{b} = \frac{1}{a} + \frac{1}{c} \][/tex]
Therefore, we have successfully shown that:
[tex]\[ \frac{3}{b} = \frac{1}{a} + \frac{1}{c} \][/tex]
