Problem 43 - Algebra 2

Question

Solve each equation. $$ \log 2 x+\log x=11 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = \sqrt{5 * 10^{10}}\) and \(x = -\sqrt{5 * 10^{10}}\).

Step 1: Combine the logarithms

Use the logarithm property \(\log_b (MN) = \log_b M + \log_b N\) to combine the two logarithms: \(\log (2x^2) = 11\)

Step 2: Eliminate the logarithm

To eliminate the logarithm, you must realize that \(\log_b M = N\) is the same as \(b^N = M\). Applying this to our problem, we obtain \(2x^2 = 10^{11}\). Divide by 2 to further simplify: \(x^2 = 5 * 10^{10}\)

Step 3: Solve for x

Take the square root of both sides to solve for x, remembering that a square root can be both positive and negative: \(x = \sqrt{5 * 10^{10}}\) and \(x = -\sqrt{5 * 10^{10}}\)

Key Concepts

Solving Logarithmic Equations

When faced with an equation involving logarithms, the main goal is to simplify and eliminate the logarithmic component to solve for the unknown variable. In our exercise

1. We start by using properties of logarithms to combine terms, reducing complexity. The original equation \( \log 2x + \log x = 11 \) is simplified to a single logarithm: \( \log (2x^2) = 11 \).
2. Next, convert the logarithmic equation into an exponential one. If \( \log_b M = N \), then \( b^N = M \). Here, assume base \( b = 10 \), giving \( 2x^2 = 10^{11} \).
3. Isolate and solve for \( x \) as you would in any algebraic equation.

Understanding how to transition from logarithmic to exponential form is key. This method allows for straightforward algebraic manipulation and access to solving for unknown variables. Remember, combining terms is the first step that guides us toward an easily manageable equation.

Properties of Logarithms

Logarithms possess several properties that make them useful in solving equations. Knowing these allows for effective manipulation of logarithmic expressions:

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

These properties mirror those of more familiar algebraic operations and provide shortcuts and simplifications in equations. Utilizing the product property in our given equation reduced the equation significantly, highlighting the power of these properties in solving complex equations efficiently. They provide a path from seemingly intricate expressions to more manageable forms, allowing for easier resolution.

Exponential Equations

An exponential equation is one in which a variable appears in the exponent. These equations often arise when solving logarithmic equations. In our solution:

• We transformed a logarithmic expression into an exponential form by recognizing that \( \log M = N \) translates numerically to \( M = 10^N \), given a base \( b \) of 10.
• This allowed us to shift from a logarithmic equation to one that could be solved using algebraic techniques.
• Exponential equations usually require using logarithms to isolate the variable initially, and then reciprocally expressing the equation as exponential for further steps.

Mastering the transition between these forms - logarithmic to exponential and vice versa - is essential in solving both types of equations. It opens a range of strategies, paving the way for precise and efficient problem solving.