Problem 4 - Algebra A Combined Function

Question

Solve each equation. Give an exact solution and a four-decimal-place approximation. $$ 3^{x}=11 $$

Step-by-Step Solution

Answer

Exact: \( x = \frac{\ln(11)}{\ln(3)} \), Approx: \( x \approx 2.1831 \).

Step 1: Understanding the Problem

We need to find the value of \( x \) such that \( 3^x = 11 \). This is an exponential equation where 11 cannot be expressed as a power of 3, so we'll use logarithms to solve for \( x \).

Step 2: Take the Logarithm of Both Sides

Apply the natural logarithm to both sides of the equation to bring the exponent down. Formally, this is \( \ln(3^x) = \ln(11) \).

Step 3: Use the Power Rule for Logarithms

Using the power rule \( \ln(a^b) = b\ln(a) \), rewrite the left side. Thus, \( x\ln(3) = \ln(11) \).

Step 4: Solve for x

Isolate \( x \) by dividing both sides by \( \ln(3) \). So, \( x = \frac{\ln(11)}{\ln(3)} \).

Step 5: Calculate the Exact Solution

The solution is already in the exact form \( x = \frac{\ln(11)}{\ln(3)} \).

Step 6: Approximate to Four Decimal Places

Calculate the value using a calculator to obtain an approximate value. Substitute the natural logarithms: \( x \approx \frac{2.3979}{1.0986} \approx 2.1831 \).

Key Concepts

Natural Logarithm

The concept of a natural logarithm, denoted as \( \ln(x) \), is a fundamental idea in mathematics. It is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828.

• The natural logarithm is used to solve exponential equations where the base is \( e \) or to bring down exponents for easier manipulation.
• This property is especially useful when the exponent cannot be expressed as a neat power of the base, as in our exercise: \( 3^x = 11 \).

Applying the natural logarithm to both sides of an equation like \( \,3^x = 11\, \), helps to draw out the exponent making it simpler to solve for \( x \). By using \( \ln(3^x) = \ln(11) \), we can then employ the power rule for logarithms to progress further.

Power Rule for Logarithms

The power rule for logarithms is a valuable tool when dealing with exponential equations. Mathematically, it is expressed as \( \ln(a^b) = b \ln(a) \).

• This rule states that the logarithm of a number raised to a power is equal to the power multiplied by the logarithm of the number.
• By using this rule, exponents can be simplified and brought down, making equations easier to solve.

In our example, after we applied the natural logarithm, we have \( \ln(3^x) = \ln(11) \). Using the power rule, this expression becomes \( x \ln(3) = \ln(11) \).This intermediate step allows us to solve for \( x \) more easily.

Exact Solution

An exact solution maintains full precision and provides the answer in its most accurate form without approximation. In our exercise, once we reached the equation \( x \ln(3) = \ln(11) \), we isolated \( x \) by dividing both sides by \( \ln(3) \).

• This gives us \( x = \frac{\ln(11)}{\ln(3)} \).
• This fraction is the exact solution, as it uses the values of the logarithms without simplifying further to a decimal.

The exact solution is important because it represents the most precise answer possible for the given problem and can be utilized for further analytical work.

Approximate Solution

While exact solutions offer precision, approximate solutions provide a practical numerical estimate. For instances where a decimal representation is preferred or required, the exact solution can be evaluated further.

• To achieve a four-decimal-place approximation for \( x = \frac{\ln(11)}{\ln(3)} \), we calculate \( \ln(11) \approx 2.3979 \) and \( \ln(3) \approx 1.0986 \).
• By computing these values via a calculator, we approximate \( x \approx \frac{2.3979}{1.0986} \approx 2.1831 \).

Hence, the approximate solution \( x \approx 2.1831 \) offers a more tangible or user-friendly form of the result, often suitable for quick calculations or when precision to every decimal is unnecessary.