If `log_(a)b=1/2,log_(b)c=1/3" and "log_(c)a=(K)/(5)`, then the value of K is

Options

A. 25
B. 35
C. 30
D. 20

Answer

To solve the problem, we need to find the value of \( K \) given the logarithmic relationships. Let's break it down step by step.

### Step 1: Write the given logarithmic equations
We are given:
1. \( \log_a b = \frac{1}{2} \)
2. \( \log_b c = \frac{1}{3} \)
3. \( \log_c a = \frac{K}{5} \)

### Step 2: Convert the logarithmic equations to their exponential forms
From the first equation:
\[ \log_a b = \frac{1}{2} \implies b = a^{\frac{1}{2}} \implies b = \sqrt{a} \]

From the second equation:
\[ \log_b c = \frac{1}{3} \implies c = b^{\frac{1}{3}} \implies c = \sqrt[3]{b} \]

### Step 3: Substitute the expression for \( b \) into the expression for \( c \)
Now substituting \( b = \sqrt{a} \) into \( c = \sqrt[3]{b} \):
\[ c = \sqrt[3]{\sqrt{a}} = a^{\frac{1}{2 \cdot 3}} = a^{\frac{1}{6}} \]

### Step 4: Find \( \log_c a \)
Now we need to find \( \log_c a \):
\[ \log_c a = \frac{\log a}{\log c} \]
We already have \( c = a^{\frac{1}{6}} \), so:
\[ \log c = \log(a^{\frac{1}{6}}) = \frac{1}{6} \log a \]

Now substituting back into the logarithmic expression:
\[ \log_c a = \frac{\log a}{\frac{1}{6} \log a} = 6 \]

### Step 5: Relate \( \log_c a \) to \( K \)
We know from the problem statement that:
\[ \log_c a = \frac{K}{5} \]
Setting this equal to our previous result:
\[ 6 = \frac{K}{5} \]

### Step 6: Solve for \( K \)
To find \( K \), we multiply both sides by 5:
\[ K = 6 \times 5 = 30 \]

Thus, the value of \( K \) is:
\[ \boxed{30} \]