Problem 40 - Algebra and Trigonometry

Question

Solve the logarithmic equation for \(x\) $$ \log _{3}(2-x)=3 $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = -25 \).

Step 1: Understand the problem

We need to solve the logarithmic equation \( \log_{3}(2-x) = 3 \) for \( x \). This means finding the value of \( x \) such that when substituted into \( 2-x \), the logarithmic equation holds true.

Step 2: Convert the logarithmic equation to an exponential equation

We use the property that if \( \log_{b}(a) = c \), then \( a = b^c \). So here, \( 2-x = 3^3 \).

Step 3: Calculate the power

Compute \( 3^3 \) which equals 27. Thus, \( 2-x = 27 \).

Step 4: Solve for x

Rearrange the equation \( 2-x = 27 \) to \( x = 2 - 27 \). This simplifies to \( x = -25 \).

Step 5: Verify the solution

Check that substituting \( x = -25 \) back into the original equation satisfies it. \( \log_{3}(2 - (-25)) = \log_{3}(27) \), which equals 3, confirming the solution is correct.

Key Concepts

Exponential Equations

Exponential equations are a type of mathematical statement where variables appear as exponents. These equations involve a base that is raised to a power. The general form is \( b^x = a \), where \( b \) is the base, \( x \) is the exponent, and \( a \) is the outcome. In many problems, the challenge is to solve for the unknown exponent.

To understand logarithmic equations better, it's helpful to learn how they relate to exponential equations. When you see an equation like \( \log_{b}(x) = c \), it can be converted into the exponential form \( x = b^c \). This relationship is crucial when dealing with logarithms, as it allows us to switch from logarithmic to exponential form, making it easier to solve.

Once in the exponential form, such as the equation in our exercise \( 2 - x = 3^3 \), the game becomes simpler. This transformation means you can directly figure out the unknown by calculating the base raised to the power and then solving for the variable as you would in a regular equation.

Properties of Logarithms

Understanding the properties of logarithms is key to solving logarithmic and many algebraic equations. Logarithms are simply the inverses of exponentials, and they present a convenient way to handle complex manipulations.

Here are some fundamental properties that are very useful:

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

These properties enable us to simplify logarithmic expressions.

In our initial problem, the logarithm property \( \log_b(a) = c \) translates to the exponential equation \( a = b^c \). This is the conversion technique we used to reframe the original equation \( \log_{3}(2-x) = 3 \) into the more workable form \( 2-x = 27 \). By applying the power property, \( 3^3 \) made it straightforward to calculate the value.

Solving for Variables

When tackling equations, one of the main goals is to solve for the unknown variable. This involves isolating the variable on one side of the equation, which may require various algebraic techniques.

For the equation \( 2-x = 27 \), the solution is found by rearranging it algebraically to isolate \( x \). Start by subtracting \( 2 \) from both sides, leading to \( -x = 25 \). Then, multiplying by \(-1\) gives \( x = -25 \).

These steps may seem simple, but maintaining balance in equations is fundamental. Every operation you perform on one side must be done to the other. This ensures the equation's integrity and leads you accurately to the solution.

Once the solution is found, it's important to verify it by substituting it back into the original equation. For instance, substituting \( x = -25 \) back into the logarithmic equation, \( \log_{3}(2-(-25)) \) indeed equals \( 3 \). This verification confirms that the solution is correct and that the interpretations and manipulations of the equation were executed properly.