If `log_2x+log_4x+log_64x=5,` find x :

Options

A. 8
B. 16
C. 7
D. 2

Answer

To solve the equation \( \log_2 x + \log_4 x + \log_{64} x = 5 \), we can follow these steps:

### Step 1: Change the bases to a common base
We can express all logarithms in terms of base 2. We know that:
- \( \log_4 x = \log_{2^2} x = \frac{1}{2} \log_2 x \)
- \( \log_{64} x = \log_{2^6} x = \frac{1}{6} \log_2 x \)

Thus, we can rewrite the equation as:
\[ \log_2 x + \frac{1}{2} \log_2 x + \frac{1}{6} \log_2 x = 5 \]

### Step 2: Combine the logarithmic terms
Let \( y = \log_2 x \). Then, we can rewrite the equation as:
\[ y + \frac{1}{2}y + \frac{1}{6}y = 5 \]

To combine these terms, we need a common denominator. The least common multiple of 1, 2, and 6 is 6. Thus, we rewrite each term:
\[ \frac{6}{6}y + \frac{3}{6}y + \frac{1}{6}y = 5 \]

Combining these gives:
\[ \frac{6y + 3y + 1y}{6} = 5 \]
\[ \frac{10y}{6} = 5 \]

### Step 3: Solve for \( y \)
Now, we can simplify:
\[ \frac{5y}{3} = 5 \]

To isolate \( y \), multiply both sides by 3:
\[ 5y = 15 \]
Now divide by 5:
\[ y = 3 \]

### Step 4: Substitute back to find \( x \)
Recall that \( y = \log_2 x \), so:
\[ \log_2 x = 3 \]

To find \( x \), we rewrite this in exponential form:
\[ x = 2^3 \]
Thus:
\[ x = 8 \]

### Final Answer
The value of \( x \) is \( 8 \).

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