Problem 111 - A Graphical Approach to College Algebra

Question

Use any method (analytic or graphical) to solve each equation. $$\log _{2} \sqrt{2 x^{2}}-1=0.5$$

Step-by-Step Solution

Answer

The solution to the equation \(\log _{2} \sqrt{2 x^{2}}-1=0.5\) is \(x = 2\).

Step 1: Simplify the Equation Inside the Logarithm

The given equation is \(\log _{2} \sqrt{2 x^{2}} -1=0.5\). First, let's simplify the term inside the logarithm. The expression \(\sqrt{2x^2}\) can be rewritten as \((2x^2)^{1/2}\). This simplifies to \(x \sqrt{2}\) for \(x \geq 0\), since the square root of \(x^2\) is \(x\). Substituting back, the equation becomes \(\log_{2} (x\sqrt{2}) - 1 = 0.5\).

Step 2: Isolate the Logarithm

Add \(1\) to both sides of the equation to isolate the logarithmic term. This gives \(\log_{2} (x\sqrt{2}) = 1.5\).

Step 3: Rewrite the Logarithmic Equation in Exponential Form

Recall that if \(\log_b A = C\), then \(b^C = A\). Applying this, we convert \(\log_{2} (x\sqrt{2}) = 1.5\) to exponential form: \((x\sqrt{2}) = 2^{1.5}\).

Step 4: Simplify the Exponential Expression

Calculate \(2^{1.5}\), which is \(2^{1+0.5} = 2^1 \times 2^{0.5} = 2 \times \sqrt{2}\). So, \(x \sqrt{2} = 2 \times \sqrt{2}\).

Step 5: Solve for x

Divide both sides by \(\sqrt{2}\) to solve for \(x\). Thus, \(x \sqrt{2} / \sqrt{2} = 2\), leading to \(x = 2\).

Step 6: Verify the Solution

Substitute \(x=2\) back into the original equation to check the correctness. Calculate: \(\log_{2} \sqrt{2(2)^2} - 1 = \log_{2} \sqrt{8} - 1 = \log_{2} 2\sqrt{2} - 1 = 0.5\), confirming the solution is correct.

Key Concepts

Logarithmic Functions

In mathematics, a logarithm is the inverse operation of an exponentiation. Logarithmic functions allow us to find out what exponent we used for a certain base to reach a wanted number. They come in handy when dealing with exponential equations or processes. Consider the expression \(y = \log_b(x)\). This tells us that \(b^y = x\). Here, the base \(b\) is typically a positive number, and often it's 10 (common logarithm) or \(e\) (natural logarithm), or any other value. In our example, we're focusing on base 2, known as the binary logarithm.

• A characteristic of logarithms is their ability to turn multiplicative processes into additive ones. This is useful in simplifying complex calculations.
• They also adhere to specific properties such as \(\log_b(m \times n) = \log_b m + \log_b n\) and \(\log_b(\frac{m}{n}) = \log_b m - \log_b n\).

Logarithmic functions can be especially useful when solving equations involving multiplication or roots, which is evident in the given exercise when simplifying \(\sqrt{2x^2}\) to \(x\sqrt{2}\). This simplification helps when solving the equation, showing the power of logarithms.

Exponential Form

Converting logarithmic equations into exponential form is a crucial step when solving them. This step often makes the solution clearer and manageable. The basic idea here is simple: when we have a logarithmic equation like \(\log_b A = C\), it implies that \(b^C = A\).

Consider our exercise: \(\log_{2} (x\sqrt{2}) = 1.5\). Another way to see this is \((x\sqrt{2}) = 2^{1.5}\). This form shows us what \(x\sqrt{2}\) equals when raised to the base of 2. It facilitates understanding of the relationship between the numbers involved.

• This exponential conversion is simple: we replace the logarithm with an equivalent exponential expression.
• Once in this form, computations such as multiplying and solving for the unknown variable become straightforward.

In this exercise, this transformation directly relates to simplifying the problem: we learn what the expression inside the logarithm equates to as an exponent, and thereby manage the solving process.

Algebraic Manipulation

Algebraic manipulation involves reorganizing equations to make them easier to solve or understand. In solving logarithmic equations, it's paramount to isolate the logarithmic expression first, which often involves moving terms around, as we did when transforming \(\log_{2} (x\sqrt{2}) - 1 = 0.5\) to \(\log_{2} (x\sqrt{2}) = 1.5\). This isolation is key in preparation for the exponential shift.

Once in exponential form, solve for the unknowns using basic algebraic operations:

• Simplifying roots and powers as seen in \(x\sqrt{2} = 2\sqrt{2}\)
• Using division to cancel terms, like when dividing by \(\sqrt{2}\) to directly solve for \(x\).

Algebraic manipulation simplifies problems considerably, allowing you to focus on solving the resulting simpler equations. In our case, it made finding \(x = 2\) straightforward by handling constants and coefficients through basic operations.

Verification of Solutions

Verification of solutions is essential in any solving process to ensure accuracy. Once we found \(x = 2\), it was important to check it against the original equation. This means substituting it back into the initial expression \(\log_{2} \sqrt{2x^2} - 1 = 0.5\).

By plugging in \(x=2\), we performed the calculations: \(\log_{2} \sqrt{8} - 1 \rightarrow \log_{2} (2\sqrt{2}) - 1\), which ultimately returns \(0.5\). This clearly corroborates the correctness of our solution.

• Verification prevents mistakes arising from overlooked simplification errors or incorrect assumptions during initial steps.
• This final check ensures that all reductions and transformations align correctly and logically solve the equation as it was posed.

In conclusion, double-checking with a substitution back into the equation successfully verifies the reliability of our derived solution.