The first term of an infinite G.R is the value of satisfying the equation `log_4(4^x-15)+ x- 2 = 0` and the common ratio is `cos(2011pi/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 follow the same reasoning as in the video transcript.

### Step 1: Solve the equation
We start with the equation given in the problem:

\[ \log_4(4^x - 15) + x - 2 = 0 \]

Rearranging gives us:

\[ \log_4(4^x - 15) = 2 - x \]

### Step 2: Take the anti-logarithm
Taking the anti-logarithm on both sides, we have:

\[ 4^x - 15 = 4^{2 - x} \]

### Step 3: Rewrite the equation
We can rewrite \(4^{2 - x}\) as:

\[ 4^2 \cdot 4^{-x} = 16 \cdot \frac{1}{4^x} \]

Thus, we have:

\[ 4^x - 15 = \frac{16}{4^x} \]

### Step 4: Let \(t = 4^x\)
Let \(t = 4^x\). The equation becomes:

\[ t - 15 = \frac{16}{t} \]

### Step 5: Multiply through by \(t\)
Multiplying through by \(t\) to eliminate the fraction gives:

\[ t^2 - 15t - 16 = 0 \]

### Step 6: Factor the quadratic equation
Now we can factor the quadratic:

\[ (t - 16)(t + 1) = 0 \]

### Step 7: Solve for \(t\)
Setting each factor to zero gives us:

\[ t - 16 = 0 \quad \Rightarrow \quad t = 16 \]
\[ t + 1 = 0 \quad \Rightarrow \quad t = -1 \quad (\text{not valid since } t = 4^x \text{ cannot be negative}) \]

Thus, we have:

\[ t = 16 \]

### Step 8: Solve for \(x\)
Since \(t = 4^x\), we have:

\[ 4^x = 16 \quad \Rightarrow \quad 4^x = 4^2 \quad \Rightarrow \quad x = 2 \]

### Step 9: Identify the first term of the G.P.
The first term \(a\) of the infinite geometric progression (G.P.) is:

\[ a = 2 \]

### Step 10: Find the common ratio
The common ratio \(r\) is given as:

\[ r = \cos\left(\frac{2011\pi}{3}\right) \]

### Step 11: Simplify the angle
To simplify \(\frac{2011\pi}{3}\):

\[ 2011 \div 6 = 335 \quad \text{(the integer part is 335, remainder is 1)} \]
Thus:

\[ \frac{2011\pi}{3} = 335 \cdot 2\pi + \frac{\pi}{3} \quad \Rightarrow \quad \cos\left(\frac{2011\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 12: Calculate the sum of the infinite G.P.
The formula for the sum \(S\) of an infinite G.P. is:

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

Substituting the values of \(a\) and \(r\):

\[ S = \frac{2}{1 - \frac{1}{2}} = \frac{2}{\frac{1}{2}} = 4 \]

### Final Answer
The sum of the infinite G.P. is:

\[ \boxed{4} \]