Problem 73 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ 15=9^{x+2} $$

Step-by-Step Solution

Answer

The exact solution in terms of logarithms is \(x=\log_9(15) - 2\); approximately, \(x\approx -0.7709\).

Step 1: Express the Equation in Logarithmic Form

To solve the equation \(15=9^{x+2}\), we need to express it in logarithmic form. Using the property of logarithms: \(a = b^c \Rightarrow \log_b(a) = c\), we write: \(x+2 = \log_9(15)\).

Step 2: Evaluate the Logarithm

To find \(x+2\), we calculate \(\log_9(15)\). Since \(\log_9(15)\) can be computed using the change of base formula, it is equivalent to \(\frac{\log_{10}(15)}{\log_{10}(9)}\).

Step 3: Calculate using a Calculator

Using a calculator, evaluate the above expression: \(\frac{\log_{10}(15)}{\log_{10}(9)} \approx 1.2291\). So, \(x+2 \approx 1.2291\).

Step 4: Solve for x

Now that we have \(x+2 \approx 1.2291\), we subtract 2 from both sides to isolate \(x\):\[x = 1.2291 - 2 = -0.7709\].

Key Concepts

Understanding Logarithmic Form

When dealing with exponential equations like the one in our exercise, it's often useful to express them in "logarithmic form". This approach can simplify the process of finding a solution.
The basic principle here is: if you have an equation in the form of \(a = b^c\), you can express it as \(\log_b(a) = c\). This transformation uses the properties of logarithms to turn what was an exponent problem into a multiplication problem. For example, in the exercise \(15 = 9^{x+2}\), we write the equation in logarithmic form as \(x + 2 = \log_9(15)\). Doing this allows us to work with the expression more comfortably, leveraging logarithmic properties to directly solve for \(x+2\). And from there, isolate \(x\) by simply rearranging the equation and performing basic arithmetic.

Exploring the Change of Base Formula

The "change of base formula" plays a crucial role when you need to compute logarithms with a base other than 10 or \(e\). The formula is:

• \(\log_b(a) = \frac{\log_k(a)}{\log_k(b)}\)

where \(k\) can be any positive number, commonly 10 (for common logarithms) or \(e\) (for natural logarithms).
This formula is extremely handy, particularly when a calculator only provides common or natural logarithmic functions. In our scenario, to calculate \(\log_9(15)\), we use this formula as follows:

• \(\log_9(15) = \frac{\log_{10}(15)}{\log_{10}(9)}\)

This allows us to calculate the value using the common logarithm, which is much easier and more universally applicable with most calculators.

Using Calculators in Math

Beginners might feel a bit apprehensive about using calculators, especially in logarithmic calculations. However, calculators are fundamental tools in math that provide speed and accuracy. To calculate an expression like \(\frac{\log_{10}(15)}{\log_{10}(9)}\), you only need a calculator capable of computing logarithms.
Most scientific calculators include a \("log"\) function, which stands for \(\log_{10}\). To solve for our expression, you'd enter:

• Enter \(\log_{10}(15)\) into the calculator and note the result.
• Then, enter \(\log_{10}(9)\) and note this result.
• Finally, divide the first result by the second result to obtain the value of \(\log_9(15)\).

By following these steps, you would find that \(\log_9(15)\) approximately equals 1.2291, which you will then use to solve for \(x\) in the final equation.