Problem 101 - College Algebra Essentials

Question

Write each equation in its equivalent exponential form. Then solve for x. $$\log _{3}(x-1)=2$$

Step-by-Step Solution

Answer

The solution to the equation \( \log_3 (x-1) = 2 \) is \( x = 10 \).

Step 1: Convert logarithmic form to exponential form

Recall that if \( \log_b a = c \), then this is equivalent to \( b^c = a \). Applying this property, the equation \( \log_3(x-1) = 2 \) is equivalent to \( 3^2 = x - 1 \).

Step 2: Simplify the equation

It is known that \( 3^2 \) equals to 9. Substituting this value into the equation, we get \( 9 = x - 1 \).

Step 3: Solve for x

Rearranging the equation \( 9 = x - 1 \) to solve for x, add 1 to both sides of the equation. This yields \( x = 9 + 1 \) which simplifies to \( x = 10 \).

Key Concepts

Converting Logarithms

Understanding how to convert logarithms to exponential form is essential when dealing with logarithmic equations. By definition, a logarithmic equation such as \( \text{log}_b(a) = c \) can be rewritten as an exponential equation where \( b^c = a \). This conversion is based on the fundamental relationship between logarithms and exponents that states the logarithm of a number is the exponent to which the base must be raised to produce that number. Let's apply this to an example.

Suppose we have \( \text{log}_3(x-1) = 2 \), to convert this logarithm to exponential form, we identify \( b = 3 \), \( a = x-1 \), and \( c = 2 \). Therefore, the exponential form is \( 3^2 = x-1 \). This step paves the way for solving the equation much more straightforwardly.

Exponential Equations

Once you have the exponential form of an equation, such as \( 3^2 = x-1 \), solving it involves basic arithmetic operations. Exponential equations like this tell us that a number (the base) raised to a power equals another number. Understanding the bases and the exponents is key to solving these types of equations. Here we notice that \( 3^2 \) is a simple calculation which results in 9. This gives us \( 9 = x - 1 \), a much more manageable equation to solve. Exponential equations often appear more complex but breaking down the terms into simpler components facilitates easier computation and further algebraic manipulation.

Solving for x

Solving for \( x \) in equations is like uncovering a mystery. The key is to isolate the variable on one side of the equation. For example, once we've determined that \( 9 = x - 1 \), we're one step away from discovering the value of \( x \). The aim is to get \( x \) alone, which in this case, simply involves adding 1 to each side of the equation. Doing so gives us \( x = 9 + 1 \) and consequently \( x = 10 \). This method of isolating \( x \) applies to various algebraic equations, facilitating that 'aha' moment when the solution is revealed.

Properties of Logarithms

Logarithms come with a set of properties that make solving logarithmic equations more manageable. The earlier mentioned conversion between logarithmic and exponential form is one of those properties. Others include the product rule, quotient rule, and power rule, which respectively allow you to log products as sums, log quotients as differences, and log powers as multiples. Understanding these properties is fundamental when manipulating and simplifying logarithmic expressions.

If we had a more complex equation, recognizing that you can turn multiplication into addition or division into subtraction using these properties could simplify the process significantly. For single logarithms, as in our example \( \text{log}_3(x-1) = 2 \), the focus is often on the conversion to exponential form, but it's helpful to be aware that these additional properties exist for more complicated equations.