- Recognize the need to isolate the exponential expression in the equation \(b^x = a\).
- Apply the common logarithm (\(\log\)) for base 10 or the natural logarithm (\(\ln\)) for other bases to both sides of the equation.
- Utilize the property \(\log(b^x) = x \cdot \log(b)\) to move the variable out of the exponent.
- Solve for \(x\) by dividing by the logarithmic constant, yielding a numerical approximation of \(\boxed{0.5635}\).
Explanation
- Understanding the Problem and Method
The problem asks us to solve an exponential equation where the solution set is expressed in terms of logarithms and evaluated as \(0.5635\). As the original equation was not explicitly stated, we will follow the standard procedure for solving exponential equations and demonstrate the process using a representative example. Exponential equations of the form \(b^x = a\) (where \(b\) is the base and \(a\) is a constant) are solved by isolating the exponential expression and then applying logarithms to both sides. By using the property that \(\log(b^x) = x \cdot \log(b)\), we can solve for \(x\) effectively. In this case, we have a decimal solution \(0.5635\), which commonly arises from equations like \(10^x = 3.66\).
- Step-by-Step Solving Procedure
To solve an equation of the form \(10^x = 3.66\), we apply the common logarithm (base 10) to both sides of the equation:
\[10^x = 3.66\]
Taking the common logarithm:
\[\log(10^x) = \log(3.66)\]
Using the logarithmic power property \(\log(b^x) = x \cdot \log(b)\) and the fact that \(\log_{10}(10) = 1\), we simplify:
\[x \cdot \log(10) = \log(3.66)\]
\[x \cdot 1 = \log(3.66)\]
\[x = \log(3.66)\]
Using a calculator to approximate the value:
\[x \approx 0.5635\]
This confirms the process for finding a numerical solution for an exponential equation.
- Final Conclusion and Solution Summary
When faced with an exponential equation involving a base other than 10 (such as \(e^x = a\)), we use the natural logarithm (\(\ln\)). For example, if we were to solve \(e^x = 1.7568\):
\[\ln(e^x) = \ln(1.7568)\]
\[x \cdot \ln(e) = \ln(1.7568)\]
\[x \cdot 1 = \ln(1.7568)\]
\[x \approx 0.5635\]
In both cases, we isolate the variable \(x\) by using the logarithmic identity \(\log_{b}(b^x) = x\). The final solution is given either in the logarithmic form (e.g., \(\log(3.66)\)) or the numerical approximation (\(0.5635\)). Since the equation itself was omitted, the process provided ensures that any similar equation can be solved systematically by identifying the base and applying the appropriate logarithm. The result, expressed numerically, is \(\boxed{0.5635}\).
Examples
In real-world scenarios, such as calculating compound interest or modeling population growth, you often encounter exponential equations. For instance, if you invest \(\$100\) at a certain interest rate and want to know how long it will take for the investment to grow to \(\$150\), you would be solving an equation like \(1.05^t = 1.5\). By applying logarithms, you can determine the time \(t\) precisely, allowing you to plan your financial goals with accuracy. This mathematical technique is a cornerstone of financial mathematics and biological growth models.