If `a^x = b^y = c^z and b/a = c/b then (2z)/(x + z)` is equal to:

Options

A. `y/x`
B. `x/y`
C. `x/2`
D. `z/x`

Answer

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

1. \( a^x = b^y = c^z \)
2. \( \frac{b}{a} = \frac{c}{b} \)

From the second equation, we can express \( b \) and \( c \) in terms of \( a \):

### Step 1: Express \( b \) and \( c \) in terms of \( a \)

Let \( b = ka \) for some constant \( k \). Then from \( \frac{b}{a} = \frac{c}{b} \), we have:

\[ \frac{ka}{a} = \frac{c}{ka} \implies k = \frac{c}{ka} \implies c = k^2 a \]

### Step 2: Substitute \( b \) and \( c \) back into the equations

Now we substitute \( b \) and \( c \) back into the first equation:

\[ a^x = (ka)^y = (k^2 a)^z \]

This gives us three equations:

1. \( a^x = k^y a^y \)
2. \( a^x = k^{2z} a^z \)

### Step 3: Equate the powers of \( a \)

Since the bases are the same, we can equate the exponents:

From \( a^x = k^y a^y \):

\[ x = y + \log_a(k^y) = y + y \log_a(k) \implies x = y(1 + \log_a(k)) \]

From \( a^x = k^{2z} a^z \):

\[ x = z + \log_a(k^{2z}) = z + 2z \log_a(k) \implies x = z(1 + 2\log_a(k)) \]

### Step 4: Set the two expressions for \( x \) equal

Now we have two expressions for \( x \):

\[ y(1 + \log_a(k)) = z(1 + 2\log_a(k)) \]

### Step 5: Solve for \( \frac{z}{x} \)

Rearranging gives us:

\[ y(1 + \log_a(k)) = z(1 + 2\log_a(k)) \]

Dividing both sides by \( z(1 + \log_a(k)) \):

\[ \frac{y}{z} = \frac{1 + 2\log_a(k)}{1 + \log_a(k)} \]

### Step 6: Find \( \frac{2z}{x + z} \)

To find \( \frac{2z}{x + z} \), we can express \( x \) in terms of \( z \):

Using \( x = z(1 + 2\log_a(k)) \):

\[ x + z = z(1 + 2\log_a(k)) + z = z(2 + 2\log_a(k)) \]

Thus,

\[ \frac{2z}{x + z} = \frac{2z}{z(2 + 2\log_a(k))} = \frac{2}{2 + 2\log_a(k)} = \frac{1}{1 + \log_a(k)} \]

### Final Result

Thus, we conclude that:

\[ \frac{2z}{x + z} = \frac{1}{1 + \log_a(k)} \]