Problem 9 - Algebra A Combined Function

Question

Solve each equation. Give an exact solution and a four-decimal-place approximation. $$ 2^{x-3}=5 $$

Step-by-Step Solution

Answer

The exact solution is \( x = 3 + \frac{\ln(5)}{\ln(2)} \), and the approximation is \( x \approx 5.3219 \).

Step 1: Identify the Equation Type

The given equation is an exponential equation because the variable \( x \) is in the exponent. The equation is \( 2^{x-3} = 5 \).

Step 2: Apply Logarithm to Both Sides

To solve an exponential equation, use a logarithm to move the variable down from the exponent. Apply the natural logarithm (ln) to both sides: \( \ln(2^{x-3}) = \ln(5) \).

Step 3: Use Logarithm Power Rule

Apply the logarithmic power rule which states \( \ln(a^b) = b \ln(a) \). This simplifies to \( (x-3) \cdot \ln(2) = \ln(5) \).

Step 4: Solve for x

Rearrange the equation to solve for \( x \). First, divide both sides by \( \ln(2) \): \( x-3 = \frac{\ln(5)}{\ln(2)} \). Then, add 3 to both sides to isolate \( x \): \( x = 3 + \frac{\ln(5)}{\ln(2)} \).

Step 5: Calculate Approximate Solution

Use a calculator to find \( \ln(5) \approx 1.6094 \) and \( \ln(2) \approx 0.6931 \). Then compute \( x = 3 + \frac{1.6094}{0.6931} \approx 5.3219 \).

Key Concepts

Understanding Exponential Equations

An exponential equation is one in which the variable appears in the exponent. These equations often involve expressions where a constant base is raised to a variable power. Our example equation is \(2^{x-3}=5\). Finding the value of \(x\) involves shifting the variable out of the exponent's position so it can be solved algebraically.
To make this shift, logarithms are commonly used, as they allow us to "bring down" the exponent. Recognizing exponential equations is the first step. Look for equations where a variable resides in the exponent, as this presence in the exponent makes the equation inherently exponential.

Utilizing Logarithms

Logarithms are mathematical tools that help us solve exponential equations by converting multiplicative relationships into additive ones. When dealing with exponential equations, applying a logarithm on both sides is a strategic move.
For example, in the equation \(2^{x-3} = 5\), applying the natural logarithm to both sides yields \(\ln(2^{x-3}) = \ln(5)\).

• This transformation allows you to utilize the power rule of logarithms, \( \ln(a^b) = b \ln(a) \), making equations more manageable by removing the variable from the exponent.
• Logarithms are invaluable in solving equations where direct algebraic manipulation isn't easily applied due to the position of the variable in the exponent.

This translation from an exponential form to a linear form is what allows us to isolate and solve for the variable.

Securing Exact Solutions

An exact solution is one that precisely satisfies the equation without approximations. For exponential equations, finding the exact value involves manipulation of the exponential properties using logarithms.
In our example \(2^{x-3} = 5\), after applying logarithms, we have \((x-3) \ln(2) = \ln(5)\). Solving for \(x\) involves isolating the variable, which results in \(x = 3 + \frac{\ln(5)}{\ln(2)}\).
Exact solutions communicate the pure mathematical relationship within the equation.

• They are expressed with operations such as addition, subtraction, multiplication, division, and the use of logarithms to show the precise algebraic manipulation.

However, since often we can't interpret these results at a glance, we may opt for practical approximate solutions.

Calculating Approximate Solutions

Approximate solutions provide a numerical value close enough to the exact solution for practical purposes. Often, calculators are used to achieve these results.
In our equation \(x = 3 + \frac{\ln(5)}{\ln(2)}\), we approximate by calculating the logarithmic values. Using a calculator, we find \(\ln(5) \approx 1.6094\) and \(\ln(2) \approx 0.6931\).
Substituting these into our equation gives \(x \approx 3 + \frac{1.6094}{0.6931} \approx 5.3219\).

• While not exact, this value is a close representation useful in practical scenarios, where a four-decimal-place accuracy suffices.
• Approximate solutions are a staple in real-world applications where precise representation isn't feasible, or when computational simplicity is prioritized.

Always check approximation methods to ensure the solution's accuracy within the context needed.