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To solve for \( x \) in the equation \( 2^x = 7 \), we can use logarithms. Here’s the step-by-step solution:

1. Start with the equation:
   \[
   2^x = 7
   \]

2. Take the logarithm of both sides (we can use either common logarithm, \( \log \), or natural logarithm, \( \ln \)):
   \[
   \ln(2^x) = \ln(7)
   \]

3. Apply the power rule of logarithms, which states \( \ln(a^b) = b \cdot \ln(a) \):
   \[
   x \cdot \ln(2) = \ln(7)
   \]

4. Solve for \( x \) by dividing both sides by \( \ln(2) \):
   \[
   x = \frac{\ln(7)}{\ln(2)}
   \]

5. Calculate the value using a calculator:
   \[
   x \approx \frac{1.9459}{0.6931} \approx 2.8074
   \]

So, the approximate irrational solution for \( x \), rounded to four decimal places, is:
\[
x \approx 2.8074
\]

Let me know if you would like more details or have any questions!

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Here are five related questions to deepen understanding:

1. What is the result if you use a different logarithmic base, such as base 10, to solve this equation?
2. How would the solution change if the equation were \( 2^x = 8 \) instead of \( 2^x = 7 \)?
3. What are the steps to check the accuracy of the approximate solution?
4. How would you solve for \( x \) if the equation were \( 3^x = 7 \) instead?
5. What is the significance of the term "irrational solution" in this context?

**Tip:** Always use logarithmic properties carefully, especially the power rule, to simplify exponential equations accurately.

Find an approximate irrational solution to 2^x = 7. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

Learn how to find an approximate irrational solution to the exponential equation 2^x = 7. This problem uses logarithmic properties and demonstrates the process of solving for x with steps. Ideal for high school students in Grades 10-12.

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