Problem 57 - Precalculus Mathematics for Calculus

Question

Solve for \(x: 2^{2 / \log _{5} x}=\frac{1}{16}\)

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{\sqrt{5}}\).

Step 1: Understand the Equation

We need to solve the equation \(2^{2/\log_{5}x} = \frac{1}{16}\). The equation involves an exponent with a fraction where the numerator is 2 and the denominator involves a logarithm.

Step 2: Rewriting the Right Side

Recognize that \(\frac{1}{16}\) can be rewritten as a power of 2. Since \(16 = 2^4\), we can write \(\frac{1}{16} = 2^{-4}\). Our equation now becomes \(2^{2/\log_{5}x} = 2^{-4}\).

Step 3: Equate the Exponents

Since the bases are the same (both are base 2), we can equate the exponents: \(\frac{2}{\log_{5}x} = -4\). This follows from the property that if \(a^m = a^n\), then \(m = n\).

Step 4: Solve for \(\log_{5}x\)

Solve the equation \(\frac{2}{\log_{5}x} = -4\) for \(\log_{5}x\). This involves cross-multiplying to get \(2 = -4 \log_{5} x\). Then, we divide both sides by \(-4\) to get \(\log_{5} x = -\frac{1}{2}\).

Step 5: Convert Logarithm to Exponential Form

Convert the logarithmic equation \(\log_{5} x = -\frac{1}{2}\) to an exponential form. This gives us \(x = 5^{-1/2}\).

Step 6: Simplify the Exponential Expression

Simplify \(x = 5^{-1/2}\) to express it in a more familiar form. Since \(5^{-1/2} = \frac{1}{\sqrt{5}}\), we find \(x = \frac{1}{\sqrt{5}}\).

Key Concepts

Logarithms

Logarithms are incredibly powerful tools in mathematics, primarily used for dealing with exponential equations. They essentially ask the question, "What power do we need to raise a certain base to in order to obtain a specific number?" In this exercise, the logarithm \(\log_{5} x\) helps manage the complexity of the exponent. When solving an exponential equation like \(2^{2 / \log _{5} x} = \frac{1}{16}\), logarithms can simplify the process and allow us to transform the equation into a more solvable form.

For instance, knowing properties of logarithms, we can convert logarithmic expressions into exponential form, aiding in solving equations. This exercise illustrates this by converting \(\log_{5} x = -\frac{1}{2}\) into \(x = 5^{-1/2}\).

Key Points About Logarithms:

• They are the inverse operations of exponentiation.
• Useful for transforming multiplication into addition, division into subtraction, and power relationships into multiplication.
• Make complex calculations more accessible by transforming exponential equations into linear form.

Solving Equations

Solving an equation involves finding the value of the variable that makes the equation true. In the exercise, we start with an equation involving exponents and a logarithm: \(2^{2 / \log _{5} x} = \frac{1}{16}\). The key is to simplify and isolate the variable.

Steps in Solving:

• First, rewrite the equation in a simpler form—transform \(\frac{1}{16}\) to \(2^{-4}\).
• Next, equate the exponents since the bases are the same. Use the property \(a^m = a^n \, \text{then} \, m = n\).
• Isolate the logarithmic expression and solve for \(\log_{5} x\).
• Convert the result from the logarithmic form to exponential form to find \(x\).

Solving equations is both about logical reasoning and applying mathematical properties to simplify an equation. Here, by using properties of exponents, the equation was broken down into manageable steps.

Properties of Exponents

Understanding the properties of exponents is crucial for simplifying and solving exponential equations. In this problem, properties of exponents help us rewrite terms and solve the equation effectively.

Key Properties:

• Equality of Exponents: If \(a^m = a^n\), then \(m = n\) if the base \(a\) is not zero. This allows us to equate the two exponents once the bases are identical, simplifying the equation to \(\frac{2}{\log_{5}x} = -4\).
• Negative Exponents: \(a^{-n} = \frac{1}{a^n}\), an essential step to convert \(\frac{1}{16}\) into \(2^{-4}\).

Using these properties allows us to transform and manipulate exponential expressions easily. They become vital tools, helping us reduce complex equations to simpler forms we can solve.