Problem 11 - A Graphical Approach to Precalculus with Limits

Question

Solve each exponential equation. Express the solution set so that (a) solutions are in exact form and, if irrational, (b) solutions are approximated to the nearest thousandth. Support your solutions by using a calculator. $$\left(\frac{1}{2}\right)^{x}=5$$

Step-by-Step Solution

Answer

Exact form: \( x \approx -2.32193 \), Approximation: \( x \approx -2.322 \).

Step 1: Write the equation in logarithmic form

We start with the equation \( \left(\frac{1}{2}\right)^{x} = 5 \). To solve for \( x \), we can take the logarithm of both sides. Using the property \( a^x = b \) leads to \( x = \log_a b \). In our case, the base is \( \frac{1}{2} \), thus, \( x = \log_{1/2} 5 \).

Step 2: Use the change of base formula

To evaluate \( \log_{1/2} 5 \) using a calculator, we apply the change of base formula: \( \log_{1/2} 5 = \frac{\log 5}{\log (1/2)} \). This allows us to calculate the logarithm using common logarithms (base 10) or natural logarithms (base \( e \)).

Step 3: Calculate logarithms using a calculator

First, calculate \( \log 5 \) and \( \log (1/2) \) separately using a calculator. You should find: \( \log 5 \approx 0.69897 \) and \( \log (1/2) \approx -0.30103 \).

Step 4: Compute the final value of x

Substitute the logarithmic values into the expression \( \frac{\log 5}{\log (1/2)} \). This results in \( x = \frac{0.69897}{-0.30103} \approx -2.32193 \).

Step 5: Approximate solution to the nearest thousandth

For an approximation to the nearest thousandth, round the value obtained: \( x \approx -2.322 \). Thus, \( x \approx -2.322 \) is the approximate solution for the equation \( \left(\frac{1}{2}\right)^{x} = 5 \).

Key Concepts

Logarithmic Form

In solving exponential equations, transforming the equation into logarithmic form is a crucial step. The original equation given is \( \left(\frac{1}{2}\right)^{x} = 5 \). When we need to solve for \( x \), converting this exponential equation to a logarithmic form simplifies finding the solution.

The rule we use is that if \( a^x = b \), then \( x = \log_a b \). The base here is \( \frac{1}{2} \), leading to the expression \( x = \log_{1/2} 5 \). By rewriting in logarithmic form, it allows us to evaluate \( x \) more efficiently using logarithmic properties and calculators.

This method provides a clearer path to determining the variable in question. You harness the power of logarithms to tackle the otherwise challenging exponents.

Change of Base Formula

The change of base formula is a handy mathematical tool used to compute logarithms with bases other than 10 or \( e \), which are the common bases available on most calculators.

For the problem \( x = \log_{1/2} 5 \), the change of base formula becomes necessary. This formula states:

• \( \log_a b = \frac{\log_c b}{\log_c a} \)

Using this, the expression transforms to \( \log_{1/2} 5 = \frac{\log 5}{\log (1/2)} \) using base 10 (common logarithms) for the calculation.

Employing the change of base formula simplifies the process, allowing for the calculation using only the base 10 or natural logarithms (\( \log \) and \( \ln \)). This method is not only straightforward but also guarantees accurate solutions when working with irrational bases.

Exact Form Solution

Arriving at an exact form solution for exponential equations means solving them without approximating until necessary. For the equation \( \left(\frac{1}{2}\right)^{x} = 5 \), we use the change of base formula to achieve an expression for \( x \) in exact form.

The solution, \( x = \frac{\log 5}{\log (1/2)} \), provides an exact form since it specifies the precise logarithmic relationship needed to solve the equation.

This expression can be evaluated for more exact numerical calculations anytime when needed. In many real-world and academic contexts, maintaining exact forms of solutions is critical to preserve information and ensure maximum accuracy. Exact form solutions are particularly valuable before any approximations are made, keeping the integrity of the mathematical process intact.

Rounding and Approximation

Once an exact solution is obtained, rounding and approximation are often applied to make the numbers more usable, especially in real-world applications where precision can sometimes be excessive.

In our example, the exact solution \( x = \frac{0.69897}{-0.30103} \) results in a calculated value of approximately \( -2.32193 \).

Rounding to the nearest thousandth leads us to \( x \approx -2.322 \). This process involves adjusting the decimal value to ensure it's succinct, often aligning with measurement or reporting standards.

Approximation balances accuracy and simplicity, a crucial practice in many fields like engineering and finance, where extreme precision could complicate problem-solving without meaningful gains.