- Isolate the exponential part by dividing the equation by 7500: \(\frac{2400}{7500} = \frac{8}{25} = 10^{-x}\).
- Apply the base-10 logarithm to both sides: \(\log\left(\frac{8}{25}\right) = \log(10^{-x})\).
- Use the power property of logarithms (\(\log(a^b) = b\log(a)\)) to simplify the right side: \(\log\left(\frac{8}{25}\right) = -x\).
- Solve for \(x\) to get the final result: \(\boxed{x = -\log \left(\frac{8}{25}\right)}\).
Explanation
- Problem Analysis
We start with the given exponential equation:
\[2400 = 7500(10)^{-x}\]
Our objective is to isolate \(x\) and find the equivalent logarithmic form from the given options. The expression involves an exponential term with base 10, which suggests that using base-10 logarithms will be the most effective strategy to solve for \(x\).
- Isolating and Simplifying the Exponential Expression
First, isolate the exponential part of the equation by dividing both sides by 7500:
\[\frac{2400}{7500} = 10^{-x}\]
We can simplify the fraction \(\frac{2400}{7500}\) by dividing both the numerator and the denominator by 300:
\[\frac{2400 \div 300}{7500 \div 300} = \frac{8}{25}\]
So the equation becomes:
\[\frac{8}{25} = 10^{-x}\]
- Applying Logarithms to Solve for \(x\)
Next, we apply the base-10 logarithm (\(\log\)) to both sides of the equation:
\[\log\left(\frac{8}{25}\right) = \log(10^{-x})\]
Using the logarithmic property \(\log(a^b) = b \cdot \log(a)\), we can simplify the right side:
\[\log\left(\frac{8}{25}\right) = -x \cdot \log(10)\]
Since the common logarithm is base 10, \(\log(10) = 1\), the equation simplifies further:
\[\log\left(\frac{8}{25}\right) = -x\]
- Conclusion
Finally, we solve for \(x\) by multiplying both sides of the equation by -1:
\[x = -\log\left(\frac{8}{25}\right)\]
Comparing this result to the given options, we find it matches the fourth option. Therefore, the equivalent logarithmic equation is \(x = -\log\left(\frac{8}{25}\right)\).
Examples
Logarithmic equations are essential for solving problems involving exponential growth or decay, such as calculating the time it takes for an investment to reach a certain value with continuous interest, or determining the time elapsed during radioactive decay. In our case, this formula could help determine how long it takes for a resource value to reduce from 7500 to 2400 at a specific decay rate. By using logarithms, we can comfortably extract variables hidden in exponents, which is a foundational skill in finance, physics, and biology.