`log_(10)x+log_(10)x^(1//2)+log_(10)x^(1//4)+....=y` and `(1+3+5+...(2y-1))/(4+7+10+....(3y+1))=(20)/(7log_(10)x)` , then x=

Options

A. `10^(5)`
B. `10^(4)`
C. `10^(3)`
D. `10^(2)`

Answer

To solve the given problem step by step, we will break it down into manageable parts.

### Step 1: Simplifying the first equation

The first equation is:

\[ \log_{10} x + \log_{10} x^{1/2} + \log_{10} x^{1/4} + \ldots = y \]

Using the property of logarithms, we can rewrite the terms:

\[ \log_{10} x^{1/2} = \frac{1}{2} \log_{10} x, \quad \log_{10} x^{1/4} = \frac{1}{4} \log_{10} x, \quad \text{and so on.} \]

Thus, we can express the series as:

\[ y = \log_{10} x \left( 1 + \frac{1}{2} + \frac{1}{4} + \ldots \right) \]

The series \(1 + \frac{1}{2} + \frac{1}{4} + \ldots\) is an infinite geometric series where the first term \(a = 1\) and the common ratio \(r = \frac{1}{2}\).

### Step 2: Finding the sum of the infinite geometric series

The sum \(S\) of an infinite geometric series can be calculated using the formula:

\[ S = \frac{a}{1 - r} \]

Substituting the values:

\[ S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

Thus, we have:

\[ y = \log_{10} x \cdot 2 = 2 \log_{10} x \]

### Step 3: Substituting \(y\) into the second equation

Now, we substitute \(y\) into the second equation:

\[ \frac{1 + 3 + 5 + \ldots + (2y - 1)}{4 + 7 + 10 + \ldots + (3y + 1)} = \frac{20}{7 \log_{10} x} \]

The numerator is the sum of the first \(y\) odd numbers, which is given by:

\[ \text{Sum of first } y \text{ odd numbers} = y^2 \]

The denominator is an arithmetic series where the first term \(a = 4\), the common difference \(d = 3\), and the number of terms is \(y\). The sum \(S\) of the first \(y\) terms of an arithmetic series is given by:

\[ S = \frac{n}{2} \cdot (2a + (n-1)d) \]

Substituting \(n = y\), \(a = 4\), and \(d = 3\):

\[ S = \frac{y}{2} \cdot (2 \cdot 4 + (y - 1) \cdot 3) = \frac{y}{2} \cdot (8 + 3y - 3) = \frac{y}{2} \cdot (3y + 5) \]

### Step 4: Setting up the equation

Now we can set up the equation:

\[ \frac{y^2}{\frac{y}{2} \cdot (3y + 5)} = \frac{20}{7 \log_{10} x} \]

Simplifying this gives:

\[ \frac{2y}{3y + 5} = \frac{20}{7 \log_{10} x} \]

### Step 5: Cross-multiplying

Cross-multiplying gives:

\[ 2y \cdot 7 \log_{10} x = 20(3y + 5) \]

This simplifies to:

\[ 14y \log_{10} x = 60y + 100 \]

### Step 6: Isolating \(\log_{10} x\)

Now, isolate \(\log_{10} x\):

\[ 14y \log_{10} x - 60y = 100 \]

Factoring out \(y\):

\[ y(14 \log_{10} x - 60) = 100 \]

Thus:

\[ 14 \log_{10} x - 60 = \frac{100}{y} \]

### Step 7: Substituting \(y = 2 \log_{10} x\)

Substituting \(y = 2 \log_{10} x\) into the equation:

\[ 14 \log_{10} x - 60 = \frac{100}{2 \log_{10} x} \]

This simplifies to:

\[ 14 \log_{10} x - 60 = \frac{50}{\log_{10} x} \]

### Step 8: Multiplying through by \(\log_{10} x\)

Multiply through by \(\log_{10} x\):

\[ 14 (\log_{10} x)^2 - 60 \log_{10} x - 50 = 0 \]

### Step 9: Solving the quadratic equation

Using the quadratic formula \(ax^2 + bx + c = 0\):

\[ a = 14, \quad b = -60, \quad c = -50 \]

The discriminant is:

\[ D = b^2 - 4ac = (-60)^2 - 4 \cdot 14 \cdot (-50) = 3600 + 2800 = 6400 \]

Thus:

\[ \log_{10} x = \frac{-(-60) \pm \sqrt{6400}}{2 \cdot 14} = \frac{60 \pm 80}{28} \]

Calculating the two possible values:

1. \(\log_{10} x = \frac{140}{28} = 5\)
2. \(\log_{10} x = \frac{-20}{28} = -\frac{5}{7}\)

### Step 10: Finding \(x\)

Thus, we have:

1. If \(\log_{10} x = 5\), then \(x = 10^5\).
2. If \(\log_{10} x = -\frac{5}{7}\), then \(x = 10^{-\frac{5}{7}}\).

### Final Answer

The value of \(x\) is:

\[ x = 10^5 \quad \text{or} \quad x = 10^{-\frac{5}{7}} \]