Problem 31 - Precalculus with Calculus Previews

Question

Solve the given logarithmic equation. $$ \log _{10} x=1+\log _{10} \sqrt{x} $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 100 \).

Step 1: Understand the Equation

We start with the equation \( \log_{10} x = 1 + \log_{10} \sqrt{x} \). Our goal is to isolate \( x \) by managing both sides of the equation.

Step 2: Use Logarithm Properties

Using the property \( \log_{10} \sqrt{x} = \frac{1}{2}\log_{10} x \), rewrite the equation: \( \log_{10} x = 1 + \frac{1}{2}\log_{10} x \).

Step 3: Move Terms to Isolate Terms involving \( x \)

Rearrange terms to have terms involving \( x \) on one side: \( \log_{10} x - \frac{1}{2}\log_{10} x = 1 \).

Step 4: Simplify the Equation

Simplify the left side: \( \frac{1}{2} \log_{10} x = 1 \).

Step 5: Solve for \( x \)

Multiply both sides by 2 to get rid of the fraction: \( \log_{10} x = 2 \).

Step 6: Exponentiate to Remove Logarithm

Exponentiate both sides: \( x = 10^2 \).

Step 7: Calculate the Final Answer

Calculate that \( x = 100 \).

Step 8: Verify the Solution

Substitute \( x = 100 \) back into the original equation to check: \( \log_{10} 100 = 1 + \log_{10} 10 \) confirms the solution since both sides equal 2.

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is essential when working with logarithmic equations. Logarithms turn multiplication into addition, division into subtraction, and exponentiation into multiplication. This characteristic makes them powerful tools for solving equations.
For instance, one useful property is that the logarithm of a square root can be rewritten. Given \( \log_b \sqrt{x} \), it can be expressed as \( \frac{1}{2} \log_b x \). This property helped in our solution when we dealt with the term \( \log_{10} \sqrt{x} \).
Remember the basic logarithmic properties:

• Product Rule: \( \log_b (mn) = \log_b m + \log_b n \)
• Quotient Rule: \( \log_b \left( \frac{m}{n} \right) = \log_b m - \log_b n \)
• Power Rule: \( \log_b (m^n) = n \log_b m \)

These rules simplify complex logarithmic expressions, making it easier to isolate variables and solve equations.

Solving Equations

Solving logarithmic equations often involves expressing both sides in terms of basic logarithmic properties. You start by managing both sides of the equation to simplify and isolate terms. This process is crucial when you have more complicated terms.
For example, in the original problem, we initially had \( \log_{10} x = 1 + \log_{10} \sqrt{x} \). By applying logarithmic properties to this equation, we simplified it to \( \log_{10} x = 1 + \frac{1}{2} \log_{10} x \).
The next step is to rearrange the equation to isolate logarithmic terms that involve the variable \( x \). Rearranging provided us with \( \frac{1}{2} \log_{10} x = 1 \).
Ultimately, solving logarithmic equations is about reducing complexity by consolidating like terms and applying algebraic principles, setting us up for the next stage: exponentiation.

Exponentiation

Exponentiation is the process of raising a number to a power. In the context of logarithmic equations, it often involves undoing a logarithm. This step transforms a logarithmic form back into its original exponential form, making it easier to solve for the variable.
In the solution, we reached the expression \( \log_{10} x = 2 \). To eliminate the logarithm, we exponentiate both sides. This means we rewrite the equation as \( x = 10^2 \). Here, we applied the rule that if \( \log_{b} a = c \), then \( a = b^c \).
Exponentiation thus allows us to find the concrete value of \( x \). Calculating \( 10^2 \) gives \( x = 100 \), providing our final answer. It’s always a good idea to verify the solution by checking the calculation back in the original equation, ensuring both sides match.