91Ó°ÊÓ

Exponential equations can often appear daunting due to involving variables in exponents. However, they can be managed by following a strategic approach.
A typical method is by substituting the conditions provided directly into the equation and solving for the unknown variables.
In the case of the function given, which is of the form \(y(x)=C e^{\alpha x}\), we have specific conditions:\( y(0) = 2 \) and \( y(1) = 1 \).
By substituting \( x = 0 \) into the equation, it simplifies to \( C = 2 \), revealing the first constant immediately.
When you substitute \( x = 1 \), the equation becomes \( 2 e^{\alpha} = 1 \), where now the task is to find \( \alpha \).

• Always start by solving for the simpler constant if possible, in our case, \( C \).
• Then use the second condition to solve for the remaining variable \( \alpha \) by isolating it or simplifying as much as possible.

Solving exponential equations often involves these initial simplifications, paving the path to using additional mathematical tools like logarithms.

Natural logarithms are fundamentally linked to exponential functions. They are incredibly useful in transforming multiplicative relationships into additive ones, making them easier to handle.
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \).
In our exercise, logarithms are employed to find \( \alpha \). When we encounter \( e^\alpha = \frac{1}{2} \), taking the natural logarithm of both sides allows us to bring down the exponent, which is key in solving for \( \alpha \).
Hence, \( \ln(e^{\alpha}) = \alpha \) and \( \ln\left(\frac{1}{2}\right) \) gives the value of \( \alpha \).

• Remember that \( \ln(e^x) = x \), making it straightforward to solve for variables in the exponent.
• The operation reverses the effect of the exponential function, essentially undoing what the exponential does.

Thus, by taking natural logarithms, an exponential problem can be broken down into an algebraic one.

Evaluating a function at specific points involves substituting given values for the variable and simplifying the expression to find the corresponding output.
After finding the constants \( C = 2 \) and \( \alpha = -\ln(2) \), the function becomes \( y(x) = 2e^{-\ln(2)x} \).
The next part of the exercise asks for \( y(2) \). This means substituting \( x = 2 \) into the simplified function. Replacing \( x \) gives us \( y(2) = 2e^{-\ln(2)\cdot 2} \).
By further simplifying this, you eventually reach \( y(2) = \frac{1}{2} \).

• Substitute the provided value into the function to determine its output at that particular point.
• Carefully simplify the expression, keeping in mind the properties of exponentials and logarithms.

Executing these steps correctly lets you accurately determine function outputs at specific points, reinforcing the understanding of both exponential and logarithmic transformations.