Problem 22 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation. Give the exact answer. $$\log _{x} 5^{2}=6$$

Step-by-Step Solution

Answer

\(x = \sqrt[6]{25}\)

Step 1: Understand the Logarithmic Equation

The given equation is \( \log_{x} 5^2 = 6 \). This means that \( x \) raised to the power of 6 equals \( 5^2 \).

Step 2: Rewrite in Exponential Form

The property of logarithms tells us that if \( \log_{b} a = c \), then \( b^c = a \). Applying that here, we rewrite the equation as \( x^6 = 5^2 \).

Step 3: Simplify the Equation

We know that \( 5^2 = 25 \). Thus, the equation becomes \( x^6 = 25 \).

Step 4: Solve the Exponential Equation

To find \( x \), take the sixth root of both sides: \( x = \sqrt[6]{25} \).

Step 5: Express the Solution

Understanding that \( \sqrt[6]{25} \) is already in its simplest radical form, our solution for \( x \) is \( \sqrt[6]{25} \).

Key Concepts

Exponential Form

The concept of exponential form is like flipping the script on a logarithm. When you have a logarithmic equation such as \( \log_{b} a = c \), it expresses that \( b \) raised to the power \( c \) equals \( a \). By understanding this, you can easily rewrite a logarithmic equation as an exponential one.
This method simplifies the process of solving equations, making it more straightforward. In the original exercise, we had \( \log_{x} 5^2 = 6 \) which cleverly translates to the exponential form \( x^6 = 25 \). This means the base \( x \) raised to the power of \( 6 \) equals \( 25 \).

• This transformation helps piece together the relationship between components in an equation.
• Recognizing this pattern can aid in solving many logarithmic equations by shifting them to an exponential viewpoint.
• It's a handy tool to understand logarithms and solve for unknowns systematically.

Properties of Logarithms

Logarithms come with a set of unique properties that make manipulating them simpler. One of these properties directly translates a logarithmic statement into an exponential form, as discussed. This principle is often the first step in solving logarithmic equations.
Knowing these properties by heart will make untangling logarithmic problems more intuitive. For example, we used the property \( \log_{b} a = c \) which implies \( b^c = a \), to switch to an easier-to-handle form.
Other useful properties include:

• Product Property: \( \log_{b}(mn) = \log_{b} m + \log_{b} n \)
• Quotient Property: \( \log_{b}(\frac{m}{n}) = \log_{b} m - \log_{b} n \)
• Power Property: \( \log_{b}(m^n) = n \cdot \log_{b} m \)

These can simplify complex logarithmic expressions or equations, making even the trickiest problems manageable.

Radical Expressions

Radical expressions involve roots, like square roots, cube roots, and beyond, forming another layer of mathematical elegance. They emerge frequently when manipulating equations, such as when you need to solve for a variable in an exponential equation.
In our exercise, the solution to \( x^6 = 25 \) necessitated the use of a sixth root to isolate \( x \). Hence, the expression \( x = \sqrt[6]{25} \) came into play.
Radicals work by undoing powers, making them essential tools in solving equations where a variable is raised to a power:

• They strip a number down to its core value needed to achieve a power when multiplied by itself a certain number of times.
• Understanding radicals is crucial when transforming exponential equations, making you adept at both simplifying and solving them.

So, whether simplifying square roots or dealing with higher roots like a sixth root, radical expressions bring the solution to life in a sometimes complicated arithmetic world.