91Ó°ÊÓ

Logarithms are an integral part of solving exponential equations. The properties of logarithms can transform complex expressions into simpler forms, making them easier to solve. Here are a few key properties:

• **Product Property**: This states that \(log_b(M imes N) = \log_b M + \log_b N\).
• **Quotient Property**: In other words, \(log_b\left(\frac{M}{N}\right)= \log_b M - \log_b N\).
• **Power Property**: This property can be expressed as \(log_b(M^k) = k \cdot \log_b M\).

In our problem, the property used is transposing from exponential form to a logarithmic form. We used the property \(b^y = x \iff \log_b{x}=y\) to change the equation from \(3^{2x} = 75\) into \(\log_3{75} = 2x\).
This becomes a straightforward algebraic equation that you can solve using basic operations.

After expressing the equation in logarithmic form, the task is now to solve for the unknown variable, which is \(x\) in this equation. Here are the steps taken in our problem:

• **Step 1:** Isolate the term containing \(x\). This is done by dividing both sides by 2, so you get \(x = \frac{\log_3{75}}{2}\).
  
• **Step 2:** Calculate the value of \(x\) using this new expression. For this, you'll need a calculator to find \(\log_3{75}\). It's approximately 4.022, leading to \(x = \frac{4.022}{2}\), which is approximately **2.011**.
  

Ensure to calculate the result up to three decimal places to comply with the exercise requirements. Always attempt to check your solutions by substituting the value back into the original equation if possible.

The change of base formula is a very handy tool when dealing with logarithms, especially when your calculator can only compute logarithms of certain bases, usually 10 or \(e\). The formula is:\(\log_b{a} = \frac{\log_c{a}}{\log_c{b}}\).
In this formula, \(c\) is any base of your choice, such as 10 for common logarithms or \(e\) for natural logarithms.

Using the change of base formula for our equation, we compute \(\log_3{75}\) using common or natural logarithms:

• \(\log_3{75} = \frac{\log{75}}{\log{3}}\) or \(\log_3{75} = \frac{\ln{75}}{\ln{3}}\)
  

This step is crucial as it allows us to handle numbers and equations even when our tools have limitations. For the case in our equation, using this method in a calculator will give you the approximate value needed to solve for \(x\). Remember, this technique is not just limited to solving equations but is also useful in various mathematical applications.