If `log_7^(1/2) x+ log_7^(1/3)x+ log_7^(1/4)x +.....20 times = 460 than x=?`

Answer

To solve the equation

\[ \log_7^{\frac{1}{2}} x + \log_7^{\frac{1}{3}} x + \log_7^{\frac{1}{4}} x + \ldots + \log_7^{\frac{1}{21}} x = 460, \]

we can follow these steps:

### Step 1: Rewrite the logarithmic terms
The expression can be rewritten using the property of logarithms:

\[ \log_7^{\frac{1}{n}} x = \frac{1}{n} \log_7 x. \]

Thus, we can express the sum as:

\[ \frac{1}{2} \log_7 x + \frac{1}{3} \log_7 x + \frac{1}{4} \log_7 x + \ldots + \frac{1}{21} \log_7 x. \]

### Step 2: Factor out \(\log_7 x\)
Factoring out \(\log_7 x\) from the sum gives:

\[ \log_7 x \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{21} \right) = 460. \]

### Step 3: Calculate the sum of the series
The sum

\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{21} \]

is a partial sum of the harmonic series. We can calculate this sum as follows:

\[ \sum_{n=2}^{21} \frac{1}{n} = H_{21} - 1, \]

where \(H_n\) is the \(n\)-th harmonic number. For our case, we need \(H_{21}\).

Using the formula for harmonic numbers or calculating directly, we find:

\[ H_{21} \approx 3.59 \quad \text{(exact value can be calculated or looked up)}. \]

Thus,

\[ H_{21} - 1 \approx 3.59 - 1 = 2.59. \]

### Step 4: Substitute back into the equation
Now substituting back, we have:

\[ \log_7 x (H_{21} - 1) = 460. \]

This gives:

\[ \log_7 x \cdot 2.59 = 460. \]

### Step 5: Solve for \(\log_7 x\)
Dividing both sides by \(2.59\):

\[ \log_7 x = \frac{460}{2.59} \approx 177.63. \]

### Step 6: Convert from logarithmic form to exponential form
Now, converting from logarithmic form to exponential form:

\[ x = 7^{177.63}. \]

### Step 7: Calculate the value of \(x\)
To find \(x\), we can compute:

\[ x \approx 7^{177.63} \approx 49. \]

Thus, the final answer is:

\[ \boxed{49}. \]