The first term of an infinite G.P is the value of satisfying the equation `log_3(3^x-8)+ x- 2 = 0` and the common ratio is `cos(22pi/3)` The sum of G.P is ?

Options

A. 1
B. `4/3`
C. 4
D. 2

Answer

To solve the problem step by step, we will break down the equation and find the first term of the infinite geometric progression (G.P.) as well as the common ratio, and then compute the sum of the G.P.

### Step 1: Solve the equation for x
We start with the equation given in the problem:
\[ \log_3(3^x - 8) + x - 2 = 0 \]

Rearranging the equation, we have:
\[ \log_3(3^x - 8) = 2 - x \]

Now, we can rewrite the logarithmic equation in exponential form:
\[ 3^{2 - x} = 3^x - 8 \]

### Step 2: Simplify the equation
This gives us:
\[ 3^{2 - x} = 3^x - 8 \]

Multiplying both sides by \(3^x\) to eliminate the logarithm:
\[ 3^2 = 3^{2x} - 8 \cdot 3^x \]
\[ 9 = 3^{2x} - 8 \cdot 3^x \]

### Step 3: Let \(y = 3^x\)
Let \(y = 3^x\). Then the equation becomes:
\[ 9 = y^2 - 8y \]

Rearranging gives us a standard quadratic equation:
\[ y^2 - 8y - 9 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -8\), and \(c = -9\):
\[ y = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} \]
\[ y = \frac{8 \pm \sqrt{64 + 36}}{2} \]
\[ y = \frac{8 \pm \sqrt{100}}{2} \]
\[ y = \frac{8 \pm 10}{2} \]

Calculating the two possible values for \(y\):
1. \(y = \frac{18}{2} = 9\)
2. \(y = \frac{-2}{2} = -1\) (not valid since \(y = 3^x\) must be positive)

Thus, we have:
\[ y = 9 \]

### Step 5: Find x
Since \(y = 3^x\), we have:
\[ 3^x = 9 \]
Taking logarithm base 3:
\[ x = 2 \]

### Step 6: Find the common ratio
The common ratio is given as:
\[ r = \cos\left(\frac{22\pi}{3}\right) \]

To simplify \(\frac{22\pi}{3}\):
\[ \frac{22\pi}{3} = 7\pi + \frac{1\pi}{3} \]
This is equivalent to:
\[ \cos\left(\frac{1\pi}{3}\right) = \cos\left(60^\circ\right) = \frac{1}{2} \]

### Step 7: Calculate the sum of the infinite G.P.
The formula for the sum of an infinite G.P. is:
\[ S_{\infty} = \frac{a}{1 - r} \]
Where \(a\) is the first term and \(r\) is the common ratio.

Substituting \(a = 2\) and \(r = \frac{1}{2}\):
\[ S_{\infty} = \frac{2}{1 - \frac{1}{2}} \]
\[ S_{\infty} = \frac{2}{\frac{1}{2}} = 4 \]

### Final Answer
The sum of the infinite G.P. is:
\[ \boxed{4} \]