Problem 36 - Algebra and Trigonometry

Question

Use the definition of the logarithmic function to find \(x\). $$ \begin{array}{ll}{\text { (a) } \log _{x} 6=\frac{1}{2}} & {\text { (b) } \log _{x} 3=\frac{1}{3}}\end{array} $$

Step-by-Step Solution

Answer

(a) \(x = 36\); (b) \(x = 27\).

Step 1: Understanding the Logarithmic Equation (a)

The equation \(\log_{x} 6 = \frac{1}{2}\) can be rewritten using the definition of logarithms as \(x^{\frac{1}{2}} = 6\). This implies that \(x\) raised to the power of \(\frac{1}{2}\) (or the square root of \(x\)) equals 6.

Step 2: Solve for x in Equation (a)

To solve the equation \(x^{\frac{1}{2}} = 6\), we need to eliminate the exponent by squaring both sides. Doing so results in \((x^{\frac{1}{2}})^2 = 6^2\). This simplifies to \(x = 36\). Thus, the value of \(x\) in this scenario is 36.

Step 3: Understanding the Logarithmic Equation (b)

The equation \(\log_{x} 3 = \frac{1}{3}\) can be rewritten, similar to step 1, as \(x^{\frac{1}{3}} = 3\). This implies that \(x\) raised to the power of \(\frac{1}{3}\) (or the cube root of \(x\)) equals 3.

Step 4: Solve for x in Equation (b)

To solve the equation \(x^{\frac{1}{3}} = 3\), we raise both sides to the power of 3 to remove the cube root, giving us \((x^{\frac{1}{3}})^3 = 3^3\). Simplifying, we find \(x = 27\). Hence, the value of \(x\) in this scenario is 27.

Key Concepts

Exponentiation

Exponentiation is a mathematical operation involving two numbers: the base and the exponent. The base tells you which number you are multiplying, and the exponent tells you how many times to use the base in the multiplication.
For example, in the expression \(x^{\frac{1}{2}}\), the base is \(x\) and the exponent is \(\frac{1}{2}\). Here, the operation signifies the square root of \(x\). This is because raising a number to the \(\frac{1}{2}\) power is equivalent to finding its square root.

• If \(x = 9\), then \(x^{\frac{1}{2}} = \sqrt{9} = 3\).
• For \(x^{\frac{1}{3}}\), we are finding the cube root. If \(x = 8\), then \(x^{\frac{1}{3}} = \sqrt[3]{8} = 2\).

Understanding exponentiation is crucial in logarithms, as it helps to unravel the logarithmic expressions such as \(\log_{x} 6 = \frac{1}{2}\) into a familiar form \(x^{\frac{1}{2}} = 6\). Without exponentiation, tackling logarithms would be much trickier.

Solving Equations

When faced with equations involving exponentiation, the key is to isolate the variable by performing operations that simplify the expression.
In the problem, the logarithmic equations are transformed into exponential form, allowing us to better visually manage and manipulate the equations.
For instance:

• Equation \(x^{\frac{1}{2}} = 6\) needs to be solved for \(x\). We square both sides to eliminate the exponent: \((x^{\frac{1}{2}})^2 = 6^2\), resulting in \(x = 36\).
• Similarly, for \(x^{\frac{1}{3}} = 3\), we cube both sides: \((x^{\frac{1}{3}})^3 = 3^3\), resulting in \(x = 27\).

These steps demonstrate that understanding the relationship between exponents and their corresponding roots is pivotal. Manipulating the structure of the equation by reversing this relationship can reveal the unknown value.

Properties of Logarithms

Logarithms are the inverse operations of exponentiation. They provide a way to solve equations where the variable is an exponent.
Here are some important properties and tips about logarithms:

• The equation \(\log_b a = c\) can be rewritten as \(b^c = a\). This is known as the exponential form of the logarithmic equation.
• Logarithms can simplify complex multiplication into addition: using \(\log(ab) = \log(a) + \log(b)\).
• Understanding logarithms requires familiarity with exponential functions. Knowing that logarithms and exponents are inverse means one can easily switch between the two to solve equations.

In solving the given problems, transforming the equations \(\log_x 6 = \frac{1}{2}\) and \(\log_x 3 = \frac{1}{3}\) into their corresponding exponential forms \(x^{\frac{1}{2}} = 6\) and \(x^{\frac{1}{3}} = 3\) allows standard methods of solving equations to apply, such as squaring or cubing both sides of the equation.