- Divide the equation by \(7500\) to isolate the exponential part: \(\frac{8}{25} = 10^{-x}\).
- Apply the common logarithm to both sides: \(\log\left(\frac{8}{25}\right) = \log\left(10^{-x}\right)\).
- Use the identity \(\log(10^{-x}) = -x\) to simplify: \(\log\left(\frac{8}{25}\right) = -x\).
- Multiply by \(-1\) to find the final expression: \(x = -\log\left(\frac{8}{25}\right)\).
- Final answer: \(\boxed{x = -\log\left(\frac{8}{25}\right)}\).
Explanation
- Analyze the given equation
We are given the exponential equation \(2400 = 7500(10)^{-x}\). Our goal is to isolate \(x\) and express it in logarithmic form by manipulating the equation using algebraic rules and logarithmic properties. The equation involves a base-10 exponent, so we will use the common logarithm (\(\log_{10}\), often written as \(\log\)) to solve for \(x\).
- Isolate the exponential term
First, we divide both sides of the equation by \(7500\) to isolate the term containing the variable \(x\):
\[\frac{2400}{7500} = 10^{-x}\]
Next, we simplify the fraction \(\frac{2400}{7500}\) by dividing both the numerator and the denominator by their greatest common divisor, which is \(300\):
\[\frac{2400 \div 300}{7500 \div 300} = \frac{8}{25}\]
So the equation becomes:
\[\frac{8}{25} = 10^{-x}\]
- Apply logarithms to solve for x
To solve for the exponent \(-x\), we apply the common logarithm (base 10) to both sides of the equation:
\[\log\left(\frac{8}{25}\right) = \log\left(10^{-x}\right)\]
Using the fundamental property of logarithms, \(\log_b(b^y) = y\), we have \(\log(10^{-x}) = -x\). Thus:
\[\log\left(\frac{8}{25}\right) = -x\]
- Conclusion and final selection
Finally, we multiply both sides of the equation by \(-1\) to isolate \(x\):
\[x = -\log\left(\frac{8}{25}\right)\]
Comparing this result with the given options, we see that it matches the third option provided:
\[x = -\log\left(\frac{8}{25}\right)\]
Examples
Logarithmic equations are used frequently in science and finance to solve for time or growth rates when the base is fixed. For example, in chemistry, pH calculations use logarithms, and in finance, the doubling time of an investment can be determined using logarithmic equations. By understanding how to rearrange these expressions, you can calculate how long it takes for an initial amount to decay to a specific value, such as the cooling of an object or the depreciation of assets.