Problem 34 - College Algebra and Calculus: An Applied Approach

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(2^{x-3}=31\)

Step-by-Step Solution

Answer

The approximate solution to the equation \(2^{x-3}=31\) is \(x \approx 7.954\).

Step 1: Convert to Logarithmic Form

The first step in solving this exponential equation is to convert it to logarithmic form. This can be done using the technique that involves the base of the exponential (2). Now the equation \(2^{x-3}=31\) can be rewritten as \(\log_{2}31=x-3\).

Step 2: Make x the Subject

After rewriting the equation in logarithmic form, the next step is to make the x the subject of the formula. This can be done by simply adding 3 to both sides of the equation to isolate x, which leads to the equation \(\log_{2}31+3=x\).

Step 3: Approximate the Result

The final step is to evaluate the logarithm on the left side of the equation and add 3 to that result. This will give an approximation for x. Using a calculator, \(\log_{2}(31)\) approximates to about 4.954, and adding 3 gives the approximate answer for x as 7.954.

Key Concepts

Logarithmic Form

Converting an exponential equation to its logarithmic form is a fundamental technique in algebra. This process involves taking an equation like \(2^{x-3}=31\) and rewriting it so that the exponent becomes the primary point of interest, turning it into \(\log_{2}31=x-3\). The \(\log\) symbol represents a logarithm, and the subscript '2' is the base of the logarithm, which corresponds to the base of the exponential expression that we started with.

Understanding the relationship between exponentiation and logarithms is key: if \(a^b=c\), then \(\log_{a}c=b\). So in essence, taking the logarithm of a number with a certain base tells us what exponent we'd need to raise that base to in order to get the original number. This operation is invaluable for solving exponential equations, as it allows us to 'bring down' the exponent and work with it directly.

Isolating Variables

Isolating variables is a technique used to solve for an unknown in an equation. It involves manipulating the equation to get the variable of interest by itself on one side of the equals sign. In the context of our problem, once we've written the equation in logarithmic form as \(\log_{2}31=x-3\), we need to isolate \(x\).

To achieve this, we perform the simple algebraic operation of adding 3 to both sides, yielding \(\log_{2}31+3=x\). This operation simplifies the equation and 'frees' the variable \(x\), making it the subject. Isolating the variable is a cornerstone of algebra because it sets the stage for us to solve the equation and find the value of the unknown quantity we're interested in.

Logarithm Evaluation

Logarithm evaluation is the process of determining the numerical value of a logarithm. Once we isolate the variable in the equation to get \(\log_{2}31+3=x\), the next step is to evaluate \(\log_{2}31\). Since this is not a base 10 or natural logarithm (which are the most common types found on calculators), we often need to use specific logarithmic rules or a calculator that can handle logarithms of any base.

Using the change of base formula, namely \(\log_{b}a = \frac{\log a}{\log b}\), where \(\log\) can be any logarithm base that your calculator supports, allows us to compute the value. Upon evaluating, you get the decimal approximation, which when added to 3 gives us the approximate value of \(x\). Such evaluations are essential to find the numeric answer to equations involving exponents described with variables.