The given equation is:

\[ 0.5 = 1 \times 10^{-12k} \]

To solve for \( k \), follow these steps:

### Step 1: Rewrite the equation
\[ 0.5 = 10^{-12k} \]

### Step 2: Take the logarithm of both sides
To solve for \( k \), take the logarithm (base 10) of both sides of the equation:

\[ \log(0.5) = \log(10^{-12k}) \]

### Step 3: Simplify using logarithm properties
\[ \log(0.5) = -12k \cdot \log(10) \]

Since \( \log(10) = 1 \):

\[ \log(0.5) = -12k \]

### Step 4: Solve for \( k \)
\[ k = \frac{\log(0.5)}{-12} \]

### Step 5: Calculate the logarithm
\[ \log(0.5) \approx -0.3010 \]

So:

\[ k = \frac{-0.3010}{-12} \]

\[ k \approx 0.0251 \]

Thus, the solution is:

\[ k \approx 0.0251 \]

Would you like any further details or have any questions?

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Here are 5 related questions:

1. What is the significance of the logarithm base in this context?
2. How does the value of \( k \) change if the base of the logarithm is different?
3. Can you solve a similar equation with a different coefficient, like \( 0.3 = 1 \times 10^{-8k} \)?
4. How does the exponential term affect the value of \( k \)?
5. What is the general approach to solving equations involving exponential functions?

**Tip:** When solving equations with exponential terms, logarithms are often the most effective tool to isolate the variable.

Discover how to solve the exponential equation 0.5 = 1 x 10^(-12k) using logarithms. This detailed solution covers key algebraic concepts, including the properties of logarithms and exponential equations. Ideal for students in Grades 10-12.

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