Problem 35 - Precalculus with Calculus Previews

Question

Solve the given logarithmic equation. $$ \log _{6} 2 x-\log _{6}(x+1)=0 $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Express as a single logarithm

The given equation is \( \log_{6} 2x - \log_{6}(x+1) = 0 \). Use the properties of logarithms to combine the expression into a single logarithm: \( \log_{6} \frac{2x}{x+1} = 0 \).

Step 2: Apply the inverse logarithm

Since \( \log_{6} \frac{2x}{x+1} = 0 \), equate the expression to \(6^0\): \( \frac{2x}{x+1} = 1 \), because the logarithm of 1 in any base is 0.

Step 3: Clear the fraction

To solve \( \frac{2x}{x+1} = 1 \), multiply both sides by \(x+1\) to eliminate the fraction: \( 2x = x + 1 \).

Step 4: Solve for x

Subtract \(x\) from both sides: \(2x - x = 1\), resulting in \(x = 1\).

Step 5: Verify the solution

Check if \( x = 1 \) meets the original equation's requirements. Substitute \( x = 1 \) back into the original equation: \( \log_{6}(2(1)) - \log_{6}(1+1) = 0 \), which simplifies to \( \log_{6}(2) - \log_{6}(2) = 0 \), confirming the equality.

Key Concepts

Properties of Logarithms

Logarithms have special properties that make them essential for solving various mathematical equations.
They allow us to rewrite expressions making it much easier to manage or solve complex equations.
This simplification is key in the problem-solving process.First, when we see a difference of two logarithms, we can use the property: **Difference to Quotient Property**

• \( \log_b(a) - \log_b(c) = \log_b \left(\frac{a}{c}\right) \)

This property is applied in our original equation, transforming it from \( \log_6 2x - \log_6(x+1) = 0 \) to \( \log_6 \left(\frac{2x}{x+1}\right) = 0 \).Here's why it works:

• Logarithms "undo" exponents, giving us roots and quotations when a subtraction is present.
• Using properties helps to simplify expressions and can often reveal hidden solutions.

In our case, the simplification showed us an algebraic equation that was easier to solve.

Inverse Logarithms

Using the concept of inverse operations is fundamental in solving logarithmic equations.
If you understand that a logarithm is the inverse of exponentiation, you can tackle more complex problems with ease.**Understanding Inverse Logarithms**

• A logarithm \( \log_b(x) \) tells you what power \( b \) must be raised to, in order to get \( x \).
• The inverse, \( b^{\log_b(x)}=x \), essentially "undoes" the logarithm.

In our solution,

• We encountered \( \log_6 \left(\frac{2x}{x+1}\right) = 0 \),
• Recognizing that \( y = \log_b(x) \) implies \( b^y = x \), we equated the expression to \(6^0\), simplifying to \(1\), since any number raised to the power of 0 is 1.

Using the inverse enabled us to eliminate the logarithm, leaving a simple algebraic equation to solve.

Solving Algebraic Equations

Once the logarithmic equation is simplified, we often face a more straightforward algebraic equation to solve.
Understanding techniques for solving these equations quickly is a critical skill.**Steps to Solve Algebraic Equations**

• Identify and simplify the equation.
• Clear any fractions or denominators.
• Isolate the variable on one side of the equation.
• Solve for the variable.

In the exercise, after applying inverse logarithms, we were left with the equation \( \frac{2x}{x+1} = 1 \).

• We multiplied both sides by \(x+1\) to clear the fraction, giving us \(2x = x + 1\).
• Then, we isolated the variable by subtracting \(x\) from both sides, resulting in \(x = 1\).

Finally, it's important to check your solution by substituting it back into the original equation for verification.
This step ensures that no errors were made during simplification or calculation, confirming that \(x = 1\) satisfies the original logarithmic equation.