If `2^(2n-3)=2048` then `(4n+3n^(2))`= _____

Options

A. 175
B. 25
C. 125
D. 75

Answer

To solve the equation \(2^{2n-3} = 2048\) and find the value of \(4n + 3n^2\), we can follow these steps:

### Step 1: Express 2048 as a power of 2
First, we need to find the prime factorization of 2048.

\[ 2048 = 2^{11} \]

### Step 2: Set the exponents equal
Since we have \(2^{2n-3} = 2^{11}\), we can set the exponents equal to each other:

\[ 2n - 3 = 11 \]

### Step 3: Solve for \(n\)
Now, we can solve for \(n\):

\[ 2n = 11 + 3 \]
\[ 2n = 14 \]
\[ n = \frac{14}{2} = 7 \]

### Step 4: Substitute \(n\) into the expression \(4n + 3n^2\)
Now that we have \(n = 7\), we can substitute this value into the expression \(4n + 3n^2\):

\[ 4n + 3n^2 = 4(7) + 3(7^2) \]

### Step 5: Calculate \(4n\) and \(3n^2\)
Calculating each term:

\[ 4(7) = 28 \]
\[ 3(7^2) = 3(49) = 147 \]

### Step 6: Add the two results
Now, we can add these two results together:

\[ 4n + 3n^2 = 28 + 147 = 175 \]

### Final Answer
Thus, the value of \(4n + 3n^2\) is:

\[ \boxed{175} \]

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