If `log_(16)x+log_(4)x+log_(2)x=14`, then x=

Options

A. `16`
B. `32`
C. `64`
D. none of these

Answer

To solve the equation \( \log_{16} x + \log_{4} x + \log_{2} x = 14 \), we can follow these steps:

### Step 1: Change the base of the logarithms
We can express all logarithms in terms of base 2.

Using the change of base formula:
\[ \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \]

We can rewrite:
\[ \log_{16} x = \frac{\log_{2} x}{\log_{2} 16} = \frac{\log_{2} x}{4} \quad (\text{since } 16 = 2^4) \]
\[ \log_{4} x = \frac{\log_{2} x}{\log_{2} 4} = \frac{\log_{2} x}{2} \quad (\text{since } 4 = 2^2) \]
\[ \log_{2} x = \log_{2} x \]

### Step 2: Substitute back into the equation
Substituting these into the original equation gives:
\[ \frac{\log_{2} x}{4} + \frac{\log_{2} x}{2} + \log_{2} x = 14 \]

### Step 3: Combine the terms
To combine the terms, we need a common denominator. The common denominator for 4, 2, and 1 is 4. Thus, we rewrite the equation as:
\[ \frac{\log_{2} x}{4} + \frac{2 \log_{2} x}{4} + \frac{4 \log_{2} x}{4} = 14 \]
This simplifies to:
\[ \frac{1 + 2 + 4}{4} \log_{2} x = 14 \]
\[ \frac{7}{4} \log_{2} x = 14 \]

### Step 4: Solve for \( \log_{2} x \)
Now, we can solve for \( \log_{2} x \):
\[ \log_{2} x = 14 \cdot \frac{4}{7} = 8 \]

### Step 5: Convert from logarithmic to exponential form
To find \( x \), we convert from logarithmic form:
\[ x = 2^{8} = 256 \]

### Final Answer
Thus, the value of \( x \) is \( 256 \).

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