Problem 66 - College Algebra with Modeling and Visualization

Question

Solve each equation. Use the change of base formula to approximate exact answers to the nearest hundredth when appropriate. $$3\left(10^{x-2}\right)=72$$

Step-by-Step Solution

Answer

The approximate solution is \(x \approx 3.38\).

Step 1: Isolate the Exponential Expression

To begin solving the equation \(3\left(10^{x-2}\right)=72\), first, you need to isolate the term involving the exponent. Do this by dividing both sides of the equation by 3. This results in:\[10^{x-2} = \frac{72}{3} = 24\]

Step 2: Apply the Change of Base Formula

To solve for \(x\), we'll need to use logarithms. We can apply the logarithm to both sides of the equation to use the change of base formula. Taking the logarithm base 10 (common logarithm) to both sides gives:\[ \log_{10}(10^{x-2}) = \log_{10}(24) \]

Step 3: Simplify Using Logarithm Laws

Utilize the power rule of logarithms, which states \( \log_b(a^n) = n \cdot \log_b(a)\). Thus, the equation becomes:\[ (x-2) \cdot \log_{10}(10) = \log_{10}(24) \]Since \( \log_{10}(10) = 1\), the equation simplifies to:\[ x-2 = \log_{10}(24) \]

Step 4: Solve for x

Solve the equation for \(x\) by adding 2 to both sides:\[ x = \log_{10}(24) + 2 \]

Step 5: Calculate the Logarithm

Use a calculator to find \( \log_{10}(24) \), which is approximately 1.38. Add 2 to this result:\[ x \approx 1.38 + 2 = 3.38 \]

Step 6: Round the Answer

Finally, round the solution to the nearest hundredth as required:\[ x \approx 3.38 \]

Key Concepts

Understanding Exponential Equations

Exponential equations are equations where the variable is in the exponent. This makes them unique and often requires specific methods to solve. In the equation \(3(10^{x-2})=72\), notice that the term \(10^{x-2}\) is the exponential expression. Exponential equations have some special features:

• They grow rapidly. For example, small increases in the exponent can lead to large changes in the overall value.
• The base of the exponent, like 10 in our example, determines the growth rate.
• To solve these equations, we typically isolate the exponential part, as seen when we divided both sides by 3 in the given problem.

Once isolated, the exponential equation can be transformed using logarithms. This is because logarithms are the inverse operations of exponentials, allowing us to solve for the variable in the exponent.

The Role of Logarithmic Functions

Logarithmic functions are the key to unlocking exponential equations. When the base of the exponent is a number we can work with, like 10, we can use common logarithms (base 10 logarithms). In the equation \(10^{x-2} = 24\), applying the logarithm to both sides assists us in moving the variable \(x\) from the exponent position. Logarithmic concepts involved here:

• Inverse property: Logarithms can reverse exponentiation. This helps transform complex equations into manageable forms.
• Change of base formula: Allows calculation of logs with bases other than 10, if necessary, using the formula \(\log_b{a} = \frac{\log_c{a}}{\log_c{b}}\).
• Logarithmic scales: They condense large ranges of data. This is why using \(\log_{10}(24)\) gives us a more digestible form to solve for \(x\).

By transforming the initial equation into \((x-2) \cdot \log_{10}(10) = \log_{10}(24)\), we utilize the logarithm to simplify progress toward finding \(x\).

Solving Equations with Logarithms

Solving exponential equations often culminates in the application of logarithms. Take for instance \((x-2) \cdot \log_{10}(10) = \log_{10}(24)\). By recognizing that \(\log_{10}(10) = 1\), we reduce the equation to \(x-2 = \log_{10}(24)\). This critical step shows how logarithms facilitate solving for the exponent in exponential equations.The process to solve for \(x\) includes:

• Simplifying the logarithmic equation by employing known logarithmic identities.
• Isolating the variable \(x\) by algebraically manipulating the simplified logarithmic expression.
• Computing the numerical value of \(\log_{10}(24)\) using a calculator for a precise solution.
• Adding or subtracting values as required by the transformed equation to isolate the variable completely, as \(x = \log_{10}(24) + 2\).

Finally, rounding the answer to a desired precision, in this exercise to the nearest hundredth, ensures clarity and usability of the resulting number, giving us \(x \approx 3.38\). Logarithms thus enable us to solve these otherwise tricky exponential equations with ease.