If `2^(n-7) xx 5^(n-4)=1250,` find the value of `n.`

Options

A. n=7
B. n=8
C. n=9
D. n=6

Answer

To solve the equation \(2^{(n-7)} \times 5^{(n-4)} = 1250\), we will follow these steps:

### Step 1: Rewrite the number 1250 in terms of its prime factors
First, we need to express 1250 as a product of its prime factors.

\[ 1250 = 125 \times 10 = 125 \times (2 \times 5) = 125 \times 2 \times 5 \]
Now, we can factor 125:
\[ 125 = 5^3 \]
So, we can rewrite 1250 as:
\[ 1250 = 5^3 \times 2^1 \]

### Step 2: Set up the equation
Now we can rewrite the original equation using the prime factorization of 1250:
\[ 2^{(n-7)} \times 5^{(n-4)} = 2^1 \times 5^3 \]

### Step 3: Equate the powers of 2 and 5
Since the bases are the same, we can equate the exponents for each base.

For the base 2:
\[ n - 7 = 1 \]

For the base 5:
\[ n - 4 = 3 \]

### Step 4: Solve for \(n\)
Now, we will solve both equations.

1. From \(n - 7 = 1\):
\[ n = 1 + 7 = 8 \]

2. From \(n - 4 = 3\):
\[ n = 3 + 4 = 7 \]

### Step 5: Check for consistency
We found two potential values for \(n\): 8 and 7. However, we need to check which one satisfies both equations.

Substituting \(n = 8\) into the second equation:
\[ 8 - 4 = 4 \quad \text{(not equal to 3)} \]

Substituting \(n = 7\) into the first equation:
\[ 7 - 7 = 0 \quad \text{(not equal to 1)} \]

Thus, the only consistent solution is:
\[ n = 8 \]

### Final Answer
The value of \(n\) is \(8\).

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