Problem 37 - Precalculus with Calculus Previews

Question

Solve the given logarithmic equation. $$ \log _{9} \sqrt{10 x+5}-\frac{1}{2}=\log _{9} \sqrt{x+1} $$

Step-by-Step Solution

Answer

x = 4

Step 1: Simplify the Equation

Let's first simplify the equation \[ \log _{9} \sqrt{10x + 5} - \frac{1}{2} = \log _{9} \sqrt{x + 1} \] by eliminating the constant term. We can add \(\frac{1}{2}\) to both sides of the equation to get \[ \log _{9} \sqrt{10x + 5} = \log _{9} \sqrt{x + 1} + \frac{1}{2} \].

Step 2: Equalize the Logarithmic Expressions

Using the properties of logarithms, if two logarithms with the same base are equal, their arguments are also equal. Therefore, we equate: \[ \sqrt{10x + 5} = \sqrt{x + 1} \times 9^{\frac{1}{2}} \].

Step 3: Simplify the Terms

Since \(9^{\frac{1}{2}} = 3\), we can simplify the equation to: \[ \sqrt{10x + 5} = 3 \cdot \sqrt{x + 1} \].

Step 4: Square Both Sides to Eliminate the Square Root

Square both sides of the equation to remove the square roots: \[ 10x + 5 = 9(x + 1) \].

Step 5: Solve for x

Now, simplify and solve for \(x\): First, expand the right side:\[ 10x + 5 = 9x + 9 \].Then, move all terms involving \(x\) to one side and constant terms to the other:\[ 10x - 9x = 9 - 5 \].This simplifies to:\[ x = 4 \].

Step 6: Verify the Solution

Substitute \( x = 4 \) back into the original logarithmic expressions to ensure both sides are equal:\( \log _{9} \sqrt{10(4) + 5} = \log _{9} \sqrt{45} \) and \( \log _{9} \sqrt{4 + 1} = \log _{9} \sqrt{5} \).These simplify to: \( \log _{9} 3\sqrt{5} = \log _{9} 3\sqrt{5} \), confirming that \( x = 4 \) is correct.

Key Concepts

Logarithmic Properties

Logarithmic properties are key tools that help us manipulate and understand equations involving logarithms. Let's break down what a logarithm actually does.
A logarithm asks the question: "To what power must the base be raised to produce a certain number?" This is expressed mathematically as \(\log_b(x) = y\), where \(b^y = x\).Knowing this, we can utilize several important properties to solve equations.

• The Product Property: \( \log_b(MN) = \log_b M + \log_b N \)
• The Quotient Property: \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \)
• The Power Property: \( \log_b(M^k) = k \log_b M \)

Understanding these properties can simplify complex equations, allowing us to break them down into more manageable parts and perform algebraic manipulations more easily.

Solving Equations

Solving equations, particularly those involving logarithms, is all about manipulation to isolate the variable of interest. Let's walk through the process:
The given equation is \[ \log _{9} \sqrt{10x + 5} - \frac{1}{2} = \log _{9} \sqrt{x + 1} \]. Our first step is to make both sides of the equation comparable. We achieve this by adding \( \frac{1}{2} \) to both sides. Doing so simplifies the equation to:
\[ \log _{9} \sqrt{10x + 5} = \log _{9} \sqrt{x + 1} + \frac{1}{2} \].
This subtraction and addition allow us to set the equations into a form where we can apply other algebraic or logarithmic manipulations. Once comparable, the next logical step is to see what other properties might allow our equation to become a standard algebraic problem.

Algebraic Manipulation

Algebraic manipulation comes into play heavily when we attempt to modify and solve equations. With the goal of isolating the variable, we need to carefully handle each term.
In this particular exercise, the equation \[ \log _{9} \sqrt{10x + 5} = \log _{9}3\sqrt{x + 1} \] really benefits from the property of equality of logarithms: \(\log_b(M) = \log_b(N) \iff M = N\).

• Taking the equation \( \sqrt{10x + 5} = 3\sqrt{x + 1} \), we square both sides to eliminate the radicals.
• This gives us: \( 10x + 5 = 9(x + 1) \).
• By distributing and simplifying, we eventually find \( x = 4\), concluding the algebraic manipulation.

Remember: Always simplify both sides throughout to maintain equality, ensuring accuracy as you progress from one step to the next.

Verification of Solutions

Verification of solutions is an essential step in solving equations, especially those involving logarithms. It ensures that the calculated variable value satisfies the original equation.
Once we've found a potential solution like \( x = 4 \), it's important to substitute it back into the original logarithmic equation to check if both sides are equal.
For example, substituting \( x = 4 \) into the logarithmic terms gives us: \[ \log _{9} \sqrt{10 \times 4 + 5} = \log _{9} 3\sqrt{5} \] and \(\log _{9} \sqrt{4 + 1} = \log _{9} \sqrt{5} \).
By simplification, both sides confirm to be equal \( \log _{9} 3 \sqrt{5} \), verifying \( x = 4 \) as correct.Always remember: Verification isn't just a double-check. It's an affirmation that your steps and understanding were correct. Failing to verify means leaving your solution open to errors, especially with potential restrictions within logarithmic domains.