Problem 86 - Precalculus

Question

Explain the mistake that is made. Solve the equation: \(\log (x)+\log (3)=1\) Solution: Apply the product property ( 5 ). \(\log (3 x)=1\) Exponentiate (base 10 ) \(10^{\log (3 x)}=1\) Apply the properties of inverses. \(3 x=1\) Solve for \(x . \quad \quad x=\frac{1}{3}\) This is incorrect. What mistake was made?

Step-by-Step Solution

Answer

The error was in solving \(3x = 1\) instead of \(3x = 10\).

Step 1: Understand the Equation

We are given the equation \( \log(x) + \log(3) = 1 \). This needs to be equivalent to a simpler expression using logarithm properties.

Step 2: Apply Logarithm Properties

Use the product property of logarithms: \( \log(a) + \log(b) = \log(ab) \). Hence, \( \log(x) + \log(3) \) becomes \( \log(3x) \), so now we have \( \log(3x) = 1 \).

Step 3: Exponentiate to Solve the Equation

Convert the logarithmic equation \( \log(3x) = 1 \) into an exponential equation: \( 3x = 10^1 \), which simplifies to \( 3x = 10 \).

Step 4: Solve for x

To find \( x \), divide both sides of the equation \( 3x = 10 \) by 3, yielding \( x = \frac{10}{3} \).

Step 5: Identify the Mistake

The mistake in the solution was in the step where they solved \( 3x = 1 \) instead of \( 3x = 10 \). They incorrectly used the equation \( 10^{\log(3x)} = 1 \) to conclude \( 3x = 1 \), neglecting that the base-10 exponent should result in \( 10^1 \), not 1.

Key Concepts

Logarithm Properties

Understanding logarithm properties is essential when solving logarithmic equations. Logarithms are the inverse operations of exponentiation. They help us find the power that a certain base must be raised to obtain a specific number. In the equation given, \[ \log(x) + \log(3) = 1 \],we aim to simplify this using logarithm properties. These properties enable us to combine or break apart logarithms to make solving equations easier.

• The primary properties of logarithms include the product property, the quotient property, and the power property.
• Each property provides a way to simplify equations by expressing complex logarithms in simpler terms.

Understanding and applying the correct properties is crucial in making the mathematics straightforward and efficient.
When you correctly apply these properties, you can tailor solutions to equations with ease.

Product Property of Logarithms

The product property of logarithms is particularly useful when you have to deal with addition inside logarithmic expressions. It states:\[ \log(a) + \log(b) = \log(ab) \]This property allows us to combine two logarithmic terms into a single logarithm. In the problem we examined, \[ \log(x) + \log(3) \]can be rewritten as:\[ \log(3x) \],thanks to the product property.

• This is the step that simplifies the addition of logarithms into a multiplicative format.
• It reduces complexity and gives rise to a simpler expression, allowing for further manipulation.

With this property, solving the equation becomes more manageable because you now deal with one logarithm instead of two separate ones. This is a powerful tool in your mathematical toolkit for simplifying and solving logarithmic equations.

Exponential Equations

Converting logarithmic equations into exponential form is a standard step in solving these types of equations. Once the equation \[ \log(3x) = 1 \]is established using the product property, we convert it into its exponential form. This transformation utilizes the definition \[ a^{\log_a(y)} = y \]to get:\[ 3x = 10^1 \].

• This step changes the problem from one of logarithms to a more straightforward algebraic equation.
• Exponential equations are often simpler to solve because they can be solved through straightforward manipulation.

By converting to exponential form, you can easily manipulate the equation to solve for unknown variables. This process is indispensable in transitioning seamlessly from logarithms to finding actual numerical solutions.