Question

Evaluate the logarithm. Round your result to three decimal places.\(\log _{1 / 3} 5\)

Short answer

The value of \( \log _{1 / 3} 5 \) rounded to three decimal places is approximately -0.631 which can be obtained by substituting the values of \( \log(5) \) and \( \log(1 / 3) \) into a calculator.

Step-by-step solution

Understand the Problem

We are asked to find the value of \( \log _{1 / 3} 5 \). This is equivalent to the question 'what exponent should I raise 1/3 to in order to get 5?' We can solve it using equivalent exponential form.

Rewrite as an equivalent exponential equation

Let's represent \( \log _{1 / 3} 5 \) as x. This means \( (1 / 3) ^x = 5 \)

Solve the equation using logarithms

We take base 10 logarithm on both sides. Using the rules of logarithms, we rewrite as \( x\log( 1 / 3) = \log(5) \)

Solve for x

Solve the equation for x by dividing both sides by \( \log(1 / 3) \): \( x = \frac{\log(5)}{\log(1 / 3)} \)

Key concepts

Logarithm Properties

Understanding the properties of logarithms is essential for solving logarithmic equations and manipulating logarithmic expressions efficiently. Logarithms, which are the inverse of exponential functions, follow unique rules that allow us to transform and compare expressions in meaningful ways.

Product Rule

One property is the product rule, which states that the logarithm of a product is equal to the sum of the logarithms of its factors: \( \log_b(mn) = \log_b(m) + \log_b(n) \).

Quotient Rule

The quotient rule tells us that the logarithm of a quotient is the difference between the logarithm of the numerator and the logarithm of the denominator: \( \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \).

Power Rule

Another important property is the power rule, which allows us to bring the exponent in an argument out in front of the logarithm: \( \log_b(m^n) = n \times \log_b(m) \).

These properties are not only helpful but also necessary for simplifying complex logarithmic expressions and for solving the equations effectively. By mastering these rules, complicated calculations can be broken down into simpler, more manageable parts.

Exponential Equations

Exponential equations feature variables in the exponent and require a specific approach for solution. Consistent with our example, recognizing the base and using logarithms to bring down the exponent is the first step.

Exponential equations can often be rewritten to have the same base, which allows us to simplify the equation by equating exponents of like bases. If we can't match the bases exactly, logarithms are invaluable tools. They enable us to take an equation like \( (1 / 3)^x = 5 \) and introduce a logarithm to both sides to extract the variable from the exponent.

Using Base 10 Logarithms

Using base 10 logarithms or natural logs (base e) are common tactics since calculators typically have these functions available. For instance, by applying the base 10 logarithm to both sides as in the step 3 of our example, the equation becomes solvable with standard algebraic methods.

The use of logarithms in such a way demonstrates the connection between exponential and logarithmic forms, showing that logarithms serve as a bridge to shift perspectives and solve otherwise daunting exponential equations.

Solving Logarithms

Solving logarithmic equations involves isolating the logarithmic term and converting the equation into exponential form or using logarithm properties to simplify. For the given problem, we started by observing \( \log_{1/3}(5) \) and deciding to denote that as \('x'\), which allowed us to set up an exponential equation.Once we have an exponential equation, if straightforward exponentiation doesn't work, we resort to taking logarithms of each side. This move can leverage the properties of logarithms to isolate the variable 'x'. In the final steps, we often find ourselves with a log ratio, something like \( x = \frac{\log(5)}{\log(1 / 3)} \), which is easily solved with a calculator.

Importance of the Base

Understanding the base of a logarithm is crucial. For example, in solving \( \log_{1/3}(5) \) as an exponent, we are really asking for the exponent to which '1/3' must be raised to yield '5'. By focusing on this key concept, we can transform and solve even the most complex logarithmic equations.

When solving logarithmic equations, it's important to watch for extraneous solutions, which are apparent solutions that don't satisfy the original equation. These can arise when we apply algebraic operations that may expand the solution set. Always check your solutions by substituting them back into the original equation.