Problem 119 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the value of the base \(b\) so that the graph of \(f(x)=\log _{b} x\) contains the given point. $$\left(\frac{1}{81}, 2\right)$$

Step-by-Step Solution

Answer

The value of the base \(b\) is \(\frac{1}{9}\).

Step 1: Insert the points into the equation

The given function is \(f(x)=\log_b x\). The point is \((\frac{1}{81}, 2)\). Let’s replace \(x\) with \(\frac{1}{81}\) and \(f(x)\) with \(2\), we obtain: \(2 = \log_b \frac{1}{81}\)

Step 2: Convert the Logarithmic Equation into Exponential Form

The logarithmic equation \(2 = \log_b \frac{1}{81}\) can be converted into exponential form according to the logarithm properties. We will get: \(b^2 = \frac{1}{81}\)

Step 3: Solve the Equation

The equation \(b^2 = \frac{1}{81}\) can be further simplified as: \(b = \sqrt{\frac{1}{81}}\). By calculating the square root of \(\frac{1}{81}\), we find that \(b = \frac{1}{9}\) or \(b = -\frac{1}{9}\)

Step 4: Discard the Negative Value

Since the base of a logarithm cannot be negative, we discard \(b = -\frac{1}{9}\) solution. Hence, \(b = \frac{1}{9}\) is the value of the base.

Key Concepts

Logarithmic to Exponential Form

Understanding how to convert logarithmic form to exponential form is a fundamental concept in algebra that can simplify the process of solving logarithmic equations. This involves rewriting a logarithmic equation, which is given by the format \( \log_b(x) = y \), into its equivalent exponential form: \( b^y = x \).

In the context of the given exercise, we start with the point \( \left(\frac{1}{81}, 2\right) \) which lies on the graph of the function \( f(x) = \log_b(x) \). According to our logarithmic equation \( 2 = \log_b\left(\frac{1}{81}\right) \), we can convert this into exponential form to get \( b^2 = \frac{1}{81} \). This conversion is crucial as it transforms the log equation into one that is often much easier to solve, using knowledge of exponents that many students find more intuitive.

Solving Logarithmic Equations

Solving logarithmic equations typically involves isolating the log term and converting the equation to exponential form, as demonstrated in the exercise. Once in exponential form, solving for the base \(b\) or the variable can proceed by using exponent rules or by simply identifying what value of \(b\) will satisfy the equation.

After converting the given logarithm to exponential form (\(b^2 = \frac{1}{81}\)), we solve for \(b\) by finding the square root of both sides. This gives us two possible solutions: \(b = \frac{1}{9}\) or \(b = -\frac{1}{9}\). However, only positive solutions are valid for the base of a logarithm. It is important to remember that the base of a logarithm must be a positive real number other than 1 for the logarithm to be defined. Therefore, we disregard \(b = -\frac{1}{9}\) and assert that \(b = \frac{1}{9}\) is our solution.

Properties of Logarithms

Logarithms have unique properties that make solving equations more manageable and understanding the relationship between exponents and logs clearer. Some key properties of logarithms include the product, quotient, and power rules, which describe how to handle multiplication, division, and exponentiation within a logarithmic expression.

Additionally, one of the most fundamental properties is that \(\log_b(b^x) = x\), which directly relates to converting logarithmic form to exponential form and vice versa. This property was implicitly used in the exercise to convert the log equation into an exponential one. Furthermore, the property that the logarithm of 1 to any base is always 0, that is, \(\log_b(1) = 0\), and the fact that the base of a logarithm must be a positive real number different from 1 because you cannot take the logarithm of a non-positive number or have a multiplicative identity as a base are all central to understanding and using logarithms in mathematical equations.