If `2^(k)-2^(k-1)=8` then the value of `k^3` is:

Options

A. 46
B. 27
C. 64
D. 8

Answer

To solve the equation \(2^k - 2^{k-1} = 8\), we can follow these steps:

### Step 1: Simplify the left side of the equation
The expression \(2^k - 2^{k-1}\) can be simplified. We can factor out \(2^{k-1}\):
\[ 2^k - 2^{k-1} = 2^{k-1}(2 - 1) = 2^{k-1}(1) = 2^{k-1} \]
So, we rewrite the equation:
\[ 2^{k-1} = 8 \]

### Step 2: Rewrite 8 as a power of 2
We know that \(8\) can be expressed as a power of \(2\):
\[ 8 = 2^3 \]
Thus, we can rewrite the equation:
\[ 2^{k-1} = 2^3 \]

### Step 3: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
\[ k - 1 = 3 \]

### Step 4: Solve for \(k\)
Now, we solve for \(k\):
\[ k = 3 + 1 = 4 \]

### Step 5: Find \(k^3\)
Now that we have the value of \(k\), we need to find \(k^3\):
\[ k^3 = 4^3 = 64 \]

Thus, the value of \(k^3\) is \(64\).

### Summary of the solution:
The value of \(k^3\) is \(64\).

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