Problem 30 - Precalculus : Building Concepts and Connections

Question

Solve the exponential equation. Round to three decimal places, when needed. $$10^{2 x^{2}+1}-8=4$$

Step-by-Step Solution

Answer

The solutions of the given equation are \(x=\sqrt{\frac{\log 12 - 1}{2}}\) and \(x=-\sqrt{\frac{\log 12 - 1}{2}}\).

Step 1: Simplify the Equation

Our given equation is \(10^{2x^2 +1} -8 = 4\). First, we need to simplify it to the form \(10^{2x^2 + 1} = 12\). This allows us to isolate the term containing the variable \(x\).

Step 2: Apply Logarithmic Conversion

Next, we need to transform the equation from its exponential form to a more manageable form using logarithms. For that, we apply the logarithm to both sides of the equation, which gives us \(\log (10^{2x^2 + 1}) = \log 12\). According to the logarithmic rule, this simplifies to \((2x^2 + 1) \log 10 = \log 12\). Since \(\log 10 = 1\), it simplifies further to \(2x^2 + 1 = \log 12\).

Step 3: Solve for X

The equation has been simplified to \(2x^2 + 1 = \log 12\). Arrange the equation to isolate \(x^2 : 2x^2 = \log 12 - 1\). Further simplification gives us \( x^2 = \frac{\log 12 - 1}{2}\). Solving this, we get two solutions for \(x\) owing to the square root, \(x=\sqrt{\frac{\log 12 - 1}{2}}\) and \(x=-\sqrt{\frac{\log 12 - 1}{2}}\).

Key Concepts

Logarithmic Conversion

When dealing with exponential equations, logarithmic conversion acts as a powerful tool to simplify and solve these equations. In essence, a logarithm helps us express the power or exponent needed to obtain a specific value. For example, in our exercise, we’re given the equation: \[10^{2x^2 + 1} = 12\] By converting this equation to its logarithmic form, we can more easily handle and solve for the unknown variable. To achieve this, we apply the logarithm to both sides of the equation. This means that our equation becomes: \[\log (10^{2x^2 + 1}) = \log 12\] Here, we have employed the logarithmic identity: \(\log (a^b) = b \log a\). This formula allows us to "bring down" the exponent in front of the logarithm. It simplifies our original equation to: \[(2x^2 + 1) \log 10 = \log 12\] Since we know \(\log 10 = 1\) (because 10 is the base of the common logarithmic system), the equation reduces to: \[2x^2 + 1 = \log 12\] This straightforward conversion from exponential to logarithmic form makes it much easier to isolate and solve for the variable involved.

Simplifying Equations

Simplifying an equation involves reducing it to its most basic form while retaining its original meaning. In our exercise, the goal is to make the exponential equation easy to handle by detaching any additional constants. Initially, we have the complex equation: \[10^{2x^2 + 1} - 8 = 4\] The first step in simplification is to clear the constant term, '-8', from the left side of the equation. We achieve this by adding 8 to both sides, which results in: \[10^{2x^2 + 1} = 12\] This step is crucial because it allows us to focus only on the term containing the unknown variable, which is a necessary process for the subsequent application of logarithmic conversion. By isolating the exponential part of the equation, we create simpler terms that can be solved more intuitively. Upon achieving this form, we have a streamlined path to solving for our variable \(x\).

Solving for Variables

Solving for variables is the final step in resolving an equation where we need to find the value of the unknown quantities. After converting and simplifying our logarithmic equation, we operate within the form: \[2x^2 + 1 = \log 12\] From here, we aim to isolate \(x^2\) by rearranging the equation: \[2x^2 = \log 12 - 1\] And then simply dividing all terms by 2: \[x^2 = \frac{\log 12 - 1}{2}\] To find the possible values of \(x\), the square root of both sides is taken. It's important to remember that squaring introduces two solutions:

• \(x = \sqrt{\frac{\log 12 - 1}{2}}\)
• \(x = -\sqrt{\frac{\log 12 - 1}{2}}\)

These solutions encompass both the positive and negative square root possibilities. By solving for \(x\) in this way, you determine all potential values for \(x\) that satisfy the original equation.