Problem 81 - Calculus for Biology and Medicine

Question

Solve for \(x\). (a) \(e^{3 x-1}=2\) (b) \(e^{-2 x}=10\) (c) \(e^{x^{2}-1}=10\)

Step-by-Step Solution

Answer

(a) x = (ln(2) + 1)/3; (b) x = -ln(10)/2; (c) x = ±√(ln(10) + 1).

Step 1: Understanding the Exponential Equation (a)

We need to solve the equation \(e^{3x - 1} = 2\). To do this, we will take the natural logarithm (ln) of both sides to remove the exponential.

Step 2: Apply Natural Logarithm to Equation (a)

Take the natural logarithm of both sides:\[\ln(e^{3x - 1}) = \ln(2)\]Using the property \(\ln(e^y) = y\), this simplifies to:\[3x - 1 = \ln(2)\]

Step 3: Solve for x in Equation (a)

Add 1 to both sides to isolate the term with \(x\):\[3x = \ln(2) + 1\]Now, divide by 3:\[x = \frac{\ln(2) + 1}{3}\]

Step 4: Understanding the Exponential Equation (b)

We need to solve the equation \(e^{-2x} = 10\). Similar to part (a), take the natural logarithm of both sides to eliminate the exponential.

Step 5: Apply Natural Logarithm to Equation (b)

Take the natural logarithm of both sides:\[\ln(e^{-2x}) = \ln(10)\]Using the property \(\ln(e^y) = y\), simplify to:\[-2x = \ln(10)\]

Step 6: Solve for x in Equation (b)

Divide both sides by -2 to solve for \(x\):\[x = -\frac{\ln(10)}{2}\]

Step 7: Understanding the Exponential Equation (c)

To solve \(e^{x^2 - 1} = 10\), we will take the natural logarithm of both sides.

Step 8: Apply Natural Logarithm to Equation (c)

Take the natural logarithm of both sides:\[\ln(e^{x^2 - 1}) = \ln(10)\]Using the property \(\ln(e^y) = y\), simplify to:\[x^2 - 1 = \ln(10)\]

Step 9: Solve for x in Equation (c)

Add 1 to both sides to isolate \(x^2\):\[x^2 = \ln(10) + 1\]Take the square root of both sides to find \(x\):\[x = \pm\sqrt{\ln(10) + 1}\]

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a powerful mathematical tool used to solve exponential equations. In many problems, especially those involving growth and decay, we need to `undo' the exponential function. This is where the natural logarithm can help. The natural logarithm is the logarithm with base \( e \). The number \( e \) is an irrational constant roughly equal to 2.718, and is widely used in mathematical calculations.

When we take the natural logarithm of an equation like \( e^{y} = a \), it simplifies to \( y = \ln(a) \). This is one of the key properties of \( \ln \): it reverses the effect of the exponential function, allowing us to isolate the variable. In our example exercises, taking the natural logarithm of both sides changed the problem from dealing with an exponential to a simpler algebraic equation, which is much easier to solve.

Solving Equations

Solving equations is all about finding the value of the variable that makes the equation true. When dealing with exponential equations, the main challenge is to remove the exponential term so that we can isolate the variable in a straightforward manner.

To accomplish this, we often take the natural logarithm of both sides of an equation. For example, in the equation \( e^{3x - 1} = 2 \), taking the \( \ln \) transforms it into \( 3x - 1 = \ln(2) \).

Once the variable is exposed, we can use basic algebraic techniques: addition, subtraction, multiplication, and division to solve for the variable. For example, in the equation \( 3x - 1 = \ln(2) \), we add 1 to both sides and then divide by 3 to find \( x \, \text{as} \, x = \frac{\ln(2) + 1}{3} \). These steps simplify exponentially challenging issues into manageable problems.

Properties of Exponents

Understanding the properties of exponents is crucial for solving equations involving exponentials. Exponential expressions have unique characteristics that can be utilized in simplifying and solving equations.

One fundamental property is that any exponential function \( e^{x} \) raised to a power \( y \) becomes \( e^{xy} \). This property is frequently used to simplify calculations or transform expressions.

Another important property is that the base \( e \) raised to a power and then inverted, as in \( e^{-x} \), becomes a fraction, \( 1/e^{x} \). This helps in simplifying the expressions by turning complex divisions into multiplications.

When solving equations, recognizing these properties allows one to identify equivalent expressions or apply logarithmic operations effectively. For instance, in the equation \( e^{-2x} = 10 \), using \( \ln \, \) simplifies it to \( -2x = \ln(10) \), allowing us to further solve for \( x \) easily. Grasping these properties empowers you to maneuver through exponential equations with greater ease.