Problem 60 - Intermediate Algebra

Question

Solve for \(x .\) See Example 3. $$ \log _{27} x=-\frac{1}{3} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{3}\).

Step 1: Understand the Logarithmic Equation

The equation given is \( \log_{27} x = -\frac{1}{3} \) which tells us that the logarithm of \(x\) with base 27 equals \(-\frac{1}{3}\). This means that 27 raised to the power of \(-\frac{1}{3}\) will give us \(x\).

Step 2: Convert Logarithmic Equation to Exponential Form

In general, the logarithmic equation \(\log_b a = c\) is equivalent to the exponential equation \(b^c = a\). Applying this, the equation becomes \(27^{-\frac{1}{3}} = x\).

Step 3: Simplify the Exponential Expression

We need to find the value of \(27^{-\frac{1}{3}}\). Recognizing that 27 is the cube of 3, we write \(27 = 3^3\). Thus, \( (3^3)^{-\frac{1}{3}} = 3^{3 \times -\frac{1}{3}} = 3^{-1} \).

Step 4: Calculate the Power of 3

Calculating \(3^{-1}\) gives us \(\frac{1}{3}\). Therefore, \(x = 3^{-1} = \frac{1}{3}\).

Key Concepts

Exponential Form

In mathematics, understanding exponential form is like having a secret key to unlocking the mysteries of logarithms. In exponential form, a number is expressed as a power, or exponent, of a base. When we see an equation like \(b^c = a\), we are using exponential form. Here, \(b\) is the base, \(c\) is the exponent, and \(a\) is the result we get after raising the base to the power of the exponent. To relate this to our problem, when we convert the logarithmic equation \(\log_{27} x = -\frac{1}{3}\) into exponential form, it becomes \(27^{-\frac{1}{3}} = x\). This tells us that 27 should be raised to the power of \(-\frac{1}{3}\) to get the value of \(x\). Knowing how to switch between these forms makes solving logarithmic equations much easier because exponential form is often simpler to manipulate and understand.

Simplifying Exponents

Simplifying exponents involves using mathematical rules to make expressions with powers easier to understand and evaluate. When working with exponents, recognize and apply properties such as multiplication and division of like bases. In our case, simplifying \(27^{-\frac{1}{3}}\) involves recognizing that 27 can be written as a power of 3 since \(27 = 3^3\). Take a closer look at how to simplify:

• First, rewrite 27 as \((3^3)\).
• Apply the rule for powers, \((a^m)^n = a^{m \times n}\), giving us \((3^3)^{-\frac{1}{3}} = 3^{3 \times -\frac{1}{3}}\).
• The calculation results in \(3^{-1}\).

By simplifying expressions thoroughly, you transform complex calculations into manageable steps. In this example, by simplifying \(27^{-\frac{1}{3}}\) to \(3^{-1}\), we prepare it for easy evaluation in the final step.

Converting Between Forms

Converting between logarithmic and exponential forms is an essential skill in mathematics, especially when dealing with exponential equations. This conversion is a critical step because it takes what might initially seem complex and turns it into something more manageable. Such conversion follows a consistent process:

• Recognize the given logarithmic equation—here, \(\log_{27} x = -\frac{1}{3}\).
• Apply the principle: if \(\log_b a = c\), then \(b^c = a\).
• Convert, acknowledging that the base \(b\) raised to the \(c\) power gives the argument \(a\), so \(27^{-\frac{1}{3}} = x\).

By becoming comfortable with these conversions, not only do you gain effectiveness in solving equations, but you also develop a deeper comprehension of the relationship between logs and exponents. This awareness opens doors to solving a wide range of mathematical problems with confidence.