Problem 56 - Precalculus : Building Concepts and Connections

Question

Use the definition of a logarithm to solve for \(x\). $$\log _{x} 9=\frac{1}{2}$$

Step-by-Step Solution

Answer

The solution of the equation \(\log _{x} 9=\frac{1}{2}\) for \(x\) is \(x=81\).

Step 1: Understand the definition of a logarithm

According to the definition of the logarithm, \(\log _{a} b\) is also equivalent to \(a^{\log _{a} b} = b\). This means the base to the power of the result is equal to the number.

Step 2: Convert the logarithmic equation into exponential form

Use the definition of the logarithm to rewrite the equation. Replace \(\log _{x} 9=\frac{1}{2}\) with \(x^{\frac{1}{2}} = 9\). Here, \(x\) is raised to the power of \(\frac{1}{2}\) to give \(9\).

Step 3: Solve the exponential equation for \(x\)

To solve for \(x\), raise both sides of the equation to the power of 2. The equation should now look like this: \(x = 9^2\).

Step 4: Evaluate the exponential expression

Calculate the value of \(9^2\) which gives \(81\). That is, \(x = 81\). From the calculations, \(x = 81\) solves the logarithmic equation.

Key Concepts

Definition of a Logarithm

Understanding the definition of a logarithm is crucial to solving logarithmic equations. A logarithm answers the question: 'To what exponent must we raise a base number to get another number?' For instance, the logarithm \( \log_b(a) \) asks for the exponent \( x \) that makes \( b^x = a \). The number \( b \) is the base, and \( a \) is the result we want after raising \( b \) to the \( x \) power. A logarithm essentially unravels an exponent, revealing what that exponent must be.

As an analogy, think of a logarithm as a detective solving an exponential mystery: if \( b^x = a \) is our case, then \( \log_b(a) \) finds the 'culprit' exponent \( x \) that makes the equation hold true. In our exercise, \( \log_{x} 9=\frac{1}{2} \) implies that we're seeking an \( x \) such that when raised to the power of \( \frac{1}{2} \) gives us 9.

Converting Logarithmic to Exponential Form

To make solving logarithmic equations manageable, conversion to exponential form is a handy strategy. The statement \( \log_b(a) = c \) is equivalent to the exponential form \( b^c = a \). This transformation leverages the mutual nature of logarithms and exponents — they are inverse functions.

Switching Perspectives

Think of converting logarithmic to exponential form as changing your viewpoint. Instead of looking for the exponent given a base and a result (logarithmic form), you proclaim the exponent and solve for the base or the result (exponential form). With the exercise \( \log _{x} 9=\frac{1}{2} \) after conversion, we declare that \( x^{\frac{1}{2}} = 9 \) and now our task is simplified: determine the value of \( x \) that satisfies this new equation.

Solving Exponential Equations

Exponential equations feature variables in the exponent, as seen with \( x^{\frac{1}{2}} = 9 \) from our exercise. Solving these equations can feel like working a puzzle: you must manipulate the equation to isolate the variable and find its value.

Exponential Extrication

To extract the variable from an exponent, we often perform operations that are inversely related to exponentiating. For instance, we can take roots or use logarithms. In the above exercise, we raise both sides to a power that will cancel the fractional exponent, in this case, squaring both sides to eliminate \( \frac{1}{2} \) and isolate \( x \) is effective. After this operation, we solve the resulting equation \( x = 9^2 \) to find that \( x = 81 \) is the solution we are seeking. The original exponential mystery is now solved, and we've determined that the value of \( x \) that makes the equation true is 81.